bash, awk, sort _{58 53} 44 bytes

Read input from stdin, one value per line

printf `awk '{print $0,length}'|sort -rnk2n`

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53 bytes: set `awk '{print $0,length($0)}'|sort -rnk2n`;echo $1
57 bytes: set `sort -n`;while((${#2}==${#1}));do shift;done;echo $1

J-uby, 22 bytes

Port of my Ruby solution.

:max_by+-[S|:+@|:-@,I]

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Arturo, 42 bytes

$=>[arrange&'x->x-10^1+size digits x|last]

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Sort by <current value> - 10^(<digits in current value> + 1), then take the last item.

Red, 98 bytes

func[b][first sort/compare b func[a b][(j: to 1 log-10 a)<(k: to 1 log-10 b)or(j = k and(a > b))]]

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Zsh, 47 bytes

set ${(n)@};(for i;(($#1==$#i))&&<<<$i)|tail -1

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JavaScript (Node.js), 50 bytes

x=>+x.map(g=t=>t?0+g(t/10^0)+t%10:'').sort().pop()

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Husk, 5 bytes

▲←ġLO

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Takes a list as an input and outputs the "the largest number with the fewest digits".

Commented:

      # Implicit input;
    O # Sort the input list
  ġL  # Group the numbers by their length = number of digits
 ←    # Take the first group, with the least digits
▲     # Take the maximal value of the group

Thunno 2 hG, 3 bytes

Ṡñl

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Thunno 2, 5 bytes

ṠñlhG

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Explanation

ṠñlhG  # Implicit input
Ṡ      # Sort the input, so the minimum item is at the start
 ñ     # Group this sorted list by the following:
  l    #  Length (for integers this is number of digits)
   h   # Get the first group (the minimum digits)
    G  # And get the maximum item in this group
       # Implicit output

Pascal (FPC), 169 167 bytes

A function which takes variable array as input:

function m(a:array of integer):integer;var i:integer;begin m:=a[0];for i in a do if(int(log10(i))<int(log10(m)))or((int(log10(i))=int(log10(m)))and(i>m))then m:=i;end;

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(Yes I know, pascal sucks at this, in addition there is no sort function in it's library..)

JavaScript (Node.js), 53 bytes

a=>a.reduce((N,n)=>~(L=Math.log10)(N)!=~L(n)^N>n?N:n)

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Explanation

For any two values N,n, we need to decide which to return, this table describes that:

         length
        | < = >
     ---+-------
value < | N n .
      = | . ? .
      > | . N n

Where . will never happen and ? means either N or n works.

This looks a lot like an exclusive or (XOR) table:

      A
    | 0 1
 ---|-----
B 0 | 0 1
  1 | 1 0

Where we XOR length(N) != length(n) with N > n. The cheapest way (I think) to get the length of a number in JS is using Math.log10, then flooring the result with a bitwise operator (in this case ~).

I also [ab]use the fact that ^ has lower operator precedence than comparisons.

C (clang), 77 bytes

m,n,i,j;f(*a,l){for(j=99;l--;m=i<j|i>j^n>m?j=i,n:m)i=log10(n=a[l]);return m;}

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Pip, 11 bytes

@Y#_AE-_SKa

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Julia, 39

f(l)=sort(l,by=x->(length("$x"),-x))[1]

ATOable

K (ngn/k), 16 15 14 bytes

|/{=/&/\#'$x}#

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• {...}# set up a filter, returning elements from the (implicit) right argument where the code in {...} returns 1s
  • #'$x count the number of characters in the string representation of each input
  • &/\ generate a two item list of the character counts and their minimum
  • =/ generate a boolean mask indicating which indices have the fewest digits
• |/ return the maximum (remaining) value

Vyxal, 5 bytes

⁽₍LN∵

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How?

⁽₍LN∵
⁽     # Next element as a lambda:
 ₍    #  Apply both of the next two elements and wrap the results in a list:
  L   #   Length...
   N  #   ...and negate
    ∵ # Get the minimum of the input by that function

Japt -h, 4 bytes

Input as an array of integer strings, output as an integer string.

üÊÎñ

Try it

üÊÎñ     :Implicit input of array
ü        :Group and sort by
 Ê       :  Length
  Î      :First element
   ñ     :Sort
         :Implicit output of last element

Factor, 25 bytes

[ all-shortest supremum ]

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Get the largest of the shortests.

Jelly, 6 bytes

DL$ÐṂṀ

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How it works

DL$ÐṂṀ - Main link. Takes a list L on the left
  $ÐṂ  - Take the elements of L for which the following is minimal:
DL     -   Digit Length
     Ṁ - Maximum of those elements

APL (Dyalog Unicode), 36 bytes

-4 thanks to @rak1507.

f←⌈/⊢×((⌊/=⊢)≢∘⍕¨)

Try it online!

Coconut, 56 bytes

This is just a Coconut-ification of the best (currently only) Python 3 answer (as well as some of the other answers), so most of the credit should go to them. Any valid Python 3 is Coconut which means it's at most the same length. Also, Coconut compiles to Python.

lambda a:sorted(sorted(a),key=lambda x:-len(str(x)))[-1] can be changed into ->sorted(sorted(_),key=->-len(str(_)))[-1]. This compiles to lambda _=None: sorted(sorted(_), key=lambda _=None: -len(str(_)))[-1] (along with a ton of other stuff like additional Coconut builtins).

Pyth, 7 3 6 bytes

eS.ml`

Test Suite

Explanation:

e      Still grab the last element
 S      Still sort
  .ml`   But prefilter the list for those with the (m)inimum length.

7 byte solution:

eSh.gl`

Test Suite

Explanation:

   .g   Group items in (implicit) input by:
     l  The length of
      ` their representation
  h     Get those with the shortest length
 S      Sort the resulting list
e       and grab the last (i.e. largest) element

Common Lisp, 136 bytes

Golfed:

(defun o(&rest r &aux(s(sort r '<)))(flet((f(x)(floor(log x 10))))(apply 'max(remove-if(lambda(x)(<(apply 'min(mapcar 'f s))(f x)))s))))

Ungolfed:

(defun o (&rest r 
      &aux (s (sort r '<)))
  (flet ((f (x) (floor (log x 10))))
    (apply 'max 
       (remove-if (lambda (x)
            (< (apply 'min (mapcar 'f s))
               (f x)))
              s))))

C++11, 109 bytes

[](auto a){sort(a.begin(),a.end(),[](auto x,auto y){return (int)log10(x)<=(int)log10(y)&&x>y;});return a[0];}

a should be a container like vector or list. E.g. (lambda as f)

cout<<f(vector<int>{ 10, 13, 19, 100, 12200 });

SQLite, 49 bytes

SELECT n FROM a ORDER BY length(n),n DESC LIMIT 1

Of course only if the list is allowed to be in a table. The type of the field should be INTEGER.

Racket 148 bytes

(λ(l)(define l2(map number->string l))(apply max(map string->number
(filter(lambda(x)(eq?(string-length x)(apply min(map string-length l2))))l2))))

Testing:

(f (list 1 5 9 12 63 102))

Output:

9

Detailed version:

(define (f sl)
  (define sl2 (map number->string sl))
  (apply max 
         (map string->number
              (filter (lambda(x)
                        (equal? (string-length x)
                                (apply min (map string-length sl2))
                                )) sl2))))

Python 2 - 41 bytes

lambda l:max((-len(`x`),x) for x in l)[1]

Java 7, 112 104 bytes

int c(int[]a){int i=a[0],j;for(int b:a)i=(j=(i+"").length()-(b+"").length())>0?b:b>i&j==0?b:i;return i;}

Different approach to save multiple bytes thanks to @Barteks2x.

Ungolfed & test cases:

Try it here.

class M{
  static int c(int[] a){
    int i = a[0],
        j;
    for(int b : a){
      i = (j = (i+"").length() - (b+"").length()) > 0
           ? b
           : b > i & j == 0
              ? b
              : i;
    }
    return i;
  }

  public static void main(String[] a){
    System.out.println(c(new int[]{ 1 }));
    System.out.println(c(new int[]{ 9 }));
    System.out.println(c(new int[]{ 1729 }));
    System.out.println(c(new int[]{ 1, 1 }));
    System.out.println(c(new int[]{ 34, 3 }));
    System.out.println(c(new int[]{ 409, 12, 13 }));
    System.out.println(c(new int[]{ 11, 11, 11, 1 }));
    System.out.println(c(new int[]{ 11, 11, 11, 11 }));
    System.out.println(c(new int[]{ 78, 99, 620, 1 }));
    System.out.println(c(new int[]{ 78, 99, 620, 100 }));
    System.out.println(c(new int[]{ 1, 5, 9, 12, 63, 102 }));
    System.out.println(c(new int[]{ 3451, 29820, 2983, 1223, 1337 }));
    System.out.println(c(new int[]{ 738, 2383, 281, 938, 212, 1010 }));
  }
}

Output:

1
9
1729
1
3
13
1
11
1
99
9
3451
938

Bash + coreutils, 58 bytes

d=`sort -n`;egrep ^.{`sed q<<<"$d"|wc -L`}$<<<"$d"|tail -1

Input format is one value per line. Golfing suggestions are welcomed.

Explanation:

d=`sort -n`                             #save the list in ascending numerical order
egrep ^.{                    }$<<<"$d"  #print only list lines having as many chars
         `sed q<<<"$d"|wc -L`                 #as the first sorted line does
|tail -1                                #and then get the last one (the answer)

C# 72 bytes

int S(int[]i){return i.OrderBy(v=>(""+v).Length).ThenBy(v=>-v).First();}

Borrowing from some of the other answers, a very efficient way to accomplish this task is to sort by number of digits, then by value descending. This moves the answer to the first position. Linq accomplishes this fairly simply with the OrderBy and ThenBy statements.

Console.WriteLine(S(new[] { 738, 2383, 281, 938, 212, 1010 }));

Output: 938

Perl 6, 18 bytes

*.min:{.chars,-$_}

Explanation:

*\        # Whatever lambda
.min:     # find the minimum using

{         # bare block lambda with implicit parameter ｢$_｣

  .chars, # number of characters first ( implicit method call on ｢$_｣ )
  -$_     # then negative of the value in case of a tie
}

Usage:

say [738, 2383, 281, 938, 212, 1010].&( *.min:{.chars,-$_} ); # 938

my &code = *.min:{.chars,-$_}

say code [78, 99, 620, 10]; # 99

C# 85 bytes

int S(int[] i){var m=i.Min();return i.Where(v=>v<m*10&&$"{v}{m}".Length%2==0).Max();}

Usage:

Console.WriteLine(S(new[] { 738, 2383, 281, 938, 212, 1010 }));

Output: 938

Explanation

Get the minimum value from the array and assume that the minimum value is the value with the shortest length.

Filter the array to remove all values greater than 10 * the minimum. i.e if the value is 8, the first check removes all values greater than 80.

Also filter to remove all values where the length of the combined string is odd. "8" + "9" is length 2 which is even, "8" + "10" is length 3 which is odd and eliminated.

int S(int[] i)
{
    var m = i.Min();
    return i.Where(v => v < m * 10 && $"{v}{m}".Length % 2 == 0).Max();
}

Mathematica, 33 31 bytes

Max@MinimalBy[#,IntegerLength]&

MinimalBy selects all the elements of the original input list with the smallest score according to IntegerLength, i.e., with the smallest number of digits; and then Max outputs the largest one.

Thanks to Martin Ender for finding, and then saving, 2 bytes for me :)

R, 72 41 36 bytes

Rewrote the function with a new approach. Golfed 5 bytes thanks to a suggestion from @bouncyball.

n=nchar(i<-scan());max(i[n==min(n)])

Explained:

        i<-scan()       # Read input from stdin
n=nchar(         );     # Count the number of characters in each number in i
max(             )      # Return the maximum of the set where
    i[n==min(n)]        # the number of characters is the minimum number of characters.

function(i){while(1){if(length(o<-i[nchar(i)==T]))return(max(o));T=T+1}}

function(i){               # Take an input i
  while(1){                # Do the following continuously:
    if(length(
        o<-i[nchar(i)==T]) # Define o to be the subset of i with numbers of length T,
      )                    # where T is 1 (a built-in!).
                           # We take the length of this subset (its size), and then pass
                           # it to if(). Thanks to weak typing, this numeric is converted
                           # to a logical value. When this occurs, zero evaluates to FALSE
                           # and any non-zero number evaluates to TRUE. Therefore, the if()
                           # is TRUE iff the subset is not empty.
      return(max(o));      # If it's true, then we just return the largest element of the
                           # subset, breaking out of our loop.
    T=T+1                  # Otherwise, increment our counter and continue.
  }
}

Retina, 24 16 bytes

O^`
O$#`
$.0
G1`

Try it online! or run all test cases.

Saved 8 bytes thanks to Martin!

The all test is using a slightly older version of the code, but the algorithm is identical. I'll update it to be closer when I get more time.

The trailing newline is significant. Sorts the numbers by reverse numeric value, then sorts them by number of digits. This leaves us with the largest number with the fewest digits in the first position, so we can just delete the remaining digits.

dc, 54 bytes

?dZsL0sN[dsNdZsL]su[dlN<u]sU[dZlL=UdZlL>ukz0<R]dsRxlNp

Explanation:

?dZsL0sN                  # read input, initialize L (length) and N (number)
[dsNdZsL]su               # macro (function) 'u' updates the values of L and N
[dlN<u]sU                 # macro 'U' calls 'u' if N < curr_nr
[dZlL=U dZlL>ukz0<R]dsR   # macro 'R' is a loop that calls 'U' if L == curr_nr_len
                          #or 'u' if L > curr_nr_len
xlNp                      # the main: call 'R' and print N at the end

Run example: 'input.txt' contains all the test cases in the question's statement

while read list;do echo "$list -> "$(dc -f program.dc <<< $list);done < input.txt

Output:

1 -> 1
9 -> 9
1729 -> 1729
1 1 -> 1
34 3 -> 3
38 39 -> 39
409 12 13 -> 13
11 11 11 1 -> 1
11 11 11 11 -> 11
78 99 620 1 -> 1
78 99 620 10 -> 99
78 99 620 100 -> 99
1 5 9 12 63 102 -> 9
3451 29820 2983 1223 1337 -> 3451
738 2383 281 938 212 1010 -> 938

Python 2, 63 30 63 bytes

lambda a:a.sort()or[`i`for i in a if len(`i`)==len(`a[0]`)][-1]

C# (85 bytes)

It's a full function:

int y(int[] t) {int m=t.Min(i=>(i+"").Length);return t.Max(n=>(n+"").Length==m?n:0);}

Unriddled:

int y(int[] t) 
{
    int m = t.Min(i => (i+"").Length);
    return t.Max(n => (n+"").Length == m ? n : 0);
}

PowerShell v2+, 41 bytes

($args[0]|sort -des|sort{"$_".length})[0]

Takes input $args, sorts it by value in -descending order (so bigger numbers are first), then sorts that by the .length in ascending order (so shorter lengths are first). We then take the [0] element, which will be the biggest number with the fewest digits.

Examples

PS C:\Tools\Scripts\golfing> @(78,99,620,1),@(78,99,620,10),@(78,99,620,100),@(1,5,9,12,63,102),@(3451,29820,2983,1223,1337),@(738,2383,281,938,212,1010)|%{($_-join',')+" -> "+(.\output-largest-number-fewest-digits.ps1 $_)}
78,99,620,1 -> 1
78,99,620,10 -> 99
78,99,620,100 -> 99
1,5,9,12,63,102 -> 9
3451,29820,2983,1223,1337 -> 3451
738,2383,281,938,212,1010 -> 938

Perl, 38 37 bytes

Includes +1 for -a

Give input on STDIN:

perl -M5.010 maxmin.pl <<< "3451 29820 2983 1223 1337"

maxmin.pl:

#!/usr/bin/perl -a
\$G[99-y///c][$_]for@F;say$#{$G[-1]}

Uses memory linear in the largest number, so don't try this on too large numbers. A solution without that flaw is 38 bytes:

#!/usr/bin/perl -p
$.++until$\=(sort/\b\S{$.}\b/g)[-1]}{

All of these are very awkward and don't feel optimal at all...

Java 8, 110 bytes

int f(int[]a){java.util.Arrays.sort(a);int i=a[0];for(int z:a)i=(z+"").length()>(i+"").length()?i:z;return i;}

Ungolfed:

int f(int[]a){
    java.util.Arrays.sort(a);                  //sorts the array in ascending order
    int i=a[0];                                //declare output variable, uses 1 less byte than using a[0]
    for(int z:a) {
        i=(z+"").length()>(i+"").length()?i:z; //for each int, if the length isn't longer set as output
    }
    return i;
}

Any elements that aren't longer in a sorted array must be a higher value.

Also lambda version (101 bytes):

a->{java.util.Arrays.sort(a);int i=a[0];for(int z:a)i=(z+"").length()>(i+"").length()?i:z;return i;};

MATL, 11 bytes

tV48\&XS0))

Input is a column vector (using ; as separator), such as

[78; 99; 620; 100]

Try it online! Or verify all test cases.

Explanation

Let's use input [78; 99; 620; 100] as an example.

t      % Input column vector implicitly. Duplicate
       %   STACK: [78; 99; 620; 100], [78; 99; 620; 100]
V      % Convert to string. Each number is a row, left-padded with spaces
       %   STACK: [78; 99; 620; 100], [' 78'; ' 99'; '620'; '100']
48\    % Modulo 48. This transforms each digit into the corresponding number,
       % and space into 32. Thus space becomes the largest "digit"
       %   STACK: [78; 99; 620; 100], [32 7 8; 32 9 9; 6 2 0; 1 0 0]
&XS    % Sort rows in lexicographical order, and push the indices of the sorting
       %   STACK: [78; 99; 620; 100], [4; 3; 1; 2]
0)     % Get last value
       %   STACK: [78; 99; 620; 100], 2
)      % Index
       %   STACK: 99
       % Implicitly display

C#, 119 bytes

int L(int[]n){var g=n.GroupBy(o=>(o+"").Length);return g.Where(o=>o.Key==g.Min(k=>k.Key)).OrderBy(o=>o).First().Max();}

Ungolfed:

public int L(int[]n)
{
  var g = n.GroupBy(o => (o+"").Length);
  return g.Where(o => o.Key == g.Min(k => k.Key)).OrderBy(o => o).First().Max();
}

Magical Linq...

Usage:

var a = new LargestNumberFewestDigits();
Console.Write(a.L(new int[] { 738, 2383, 281, 938, 212, 1010 }));

Output:

938

J, 21 14 bytes

Saved 7 bytes thanks to miles and (indirectly) Jonathan!

{.@/:#@":"0,.-

This is a four-chain:

{.@/: (#@":"0 ,. -)

Let's walk over the input 10 27 232 1000. The inner fork consists of three tines. #@":"0 calculates the sizes, ,. concats each size with its negated (-) member. For input 10 27 232 1000, we are left with this:

   (#@":"0 ,. -) 10 27 232 1000
2   _10
2   _27
3  _232
4 _1000

Now, we have {.@/: as the outer tine. This is monadic first ({.) over dyadic sort (/:). That is, we'll be taking the first element of the result of dyadic /:. This sorts its right argument according to its left argument, which gives us for our input:

   (/: #@":"0 ,. -) 10 27 232 1000
27 10 232 1000

Then, using {. gives us the first element of that list, and we are done:

   ({.@/: #@":"0 ,. -) 10 27 232 1000
27

Old version

>./@(#~]=<./@])#@":"0

Still working on improvements. I golfed it down from 30, and I think this is good enough. I'm going to first break it down into basic parts:

   size =: #@":"0
   max =: >./
   min =: <./
   over =: @
   right =: ]
   left =: [
   selectMin =: #~ right = min over right

   f =: max over selectMin size
   f 3 4 5
5
   f 3 4 53
4
   f 343 42 53
53

Here's how this works.

>./@(#~ ] = <./@]) #@":"0

This is a monadic train, but this part is a hook. The verb >./@(#~ ] = <./@]) is called with left argument as the input to the main chain and the sizes, defined as #@":"0, as the right argument. This is computed as length (#) over (@) default format (":), that is, numeric stringification, which is made to apply to the 0-cells (i.e. members) of the input ("0).

Let's walk over the example input 409 12 13.

   (#@":"0) 409 12 13
3 2 2

Now for the inner verb, >./@(#~ ] = <./@]). It looks like >./@(...), which effectively means maximum value (>./) of (@) what's inside (...). As for the inside, this is a four-train, equivalent to this five-train:

[ #~ ] = <./@]

[ refers to the original argument, and ] refers to the size array; 409 12 13 and 3 2 2 respectively in this example. The right tine, <./@], computes the minimum size, 2 in this case. ] = <./@] is a boolean array of values equal to the minimum, 0 1 1 in this case. Finally, [ #~ ... takes values from the left argument according the right-argument mask. This means that elements that correspond to 0 are dropped and 1 retained. So we are left with 12 13. Finally, according to the above, the max is taken, giving us the correct result of 13, and we are done.

Pyke, 7 bytes

.#_il)h

Try it here!

PHP , 86 Bytes

<?$l=strlen($r=min($a=$_GET[a]));foreach($a as$v)if($v>$r&strlen($v)==$l)$r=$v;echo$r;

JavaScript (ES6), 51

l=>l.sort((a,b)=>(a+l).length-(b+l).length||b-a)[0]

Test

f=l=>l.sort((a,b)=>(a+l).length-(b+l).length||b-a)[0]

;[
 [[1], 1]
,[[9], 9]
,[[1729], 1729]
,[[1, 1], 1]
,[[34, 3], 3]
,[[38, 39], 39]
,[[409, 12, 13], 13]
,[[11, 11, 11, 1], 1]
,[[11, 11, 11, 11], 11]
,[[78, 99, 620, 1], 1]
,[[78, 99, 620, 10], 99]
,[[78, 99, 620, 100], 99]
,[[1, 5, 9, 12, 63, 102], 9]
,[[3451, 29820, 2983, 1223, 1337], 3451]
,[[738, 2383, 281, 938, 212, 1010], 938]
].forEach(([l,x])=>{
  var r=f(l)
  console.log(r==x?'OK':'KO',l+' -> '+r)
})  

Javascript (ES6), 57 54 53 bytes

l=>l.sort((a,b)=>(s=a=>1/a+`${a}`.length)(a)-s(b))[0]

For the record, my previous version was more math-oriented but 1 byte bigger:

l=>l.sort((a,b)=>(s=a=>1/a-~Math.log10(a))(a)-s(b))[0]

Test cases

let f =
l=>l.sort((a,b)=>(s=a=>1/a+`${a}`.length)(a)-s(b))[0]

console.log(f([1]));                              //  -> 1
console.log(f([9]));                              //  -> 9
console.log(f([1729]));                           //  -> 1729
console.log(f([1, 1]));                           //  -> 1
console.log(f([34, 3]));                          //  -> 3
console.log(f([38, 39]));                         //  -> 39
console.log(f([409, 12, 13]));                    //  -> 13
console.log(f([11, 11, 11, 1]));                  //  -> 1
console.log(f([11, 11, 11, 11]));                 //  -> 11
console.log(f([78, 99, 620, 1]));                 //  -> 1
console.log(f([78, 99, 620, 10]));                //  -> 99
console.log(f([78, 99, 620, 100]));               //  -> 99
console.log(f([1, 5, 9, 12, 63, 102]));           //  -> 9
console.log(f([3451, 29820, 2983, 1223, 1337]));  //  -> 3451
console.log(f([738, 2383, 281, 938, 212, 1010])); //  -> 938

Haskell, 39 bytes

snd.maximum.map((0-).length.show>>=(,))

05AB1E, 5 bytes

Code:

({é¬(

Explanation:

(      # Negate the list, e.g. [22, 33, 4] -> [-22, -33, -4]
 {     # Sort, e.g. [-22, -33, -4] -> [-33, -22, -4]
  é    # Sort by length, e.g. [-33, -22, -4] -> [-4, -22, -33]
   ¬   # Get the first element.
    (  # And negate that.

Uses the CP-1252 encoding. Try it online!

Brachylog, 16 bytes

or:@]feL:la#=,Lh

Try it online!

Explanation

or                 Sort the list in descending order.
  :@]f             Find all suffixes of the list.
      eL           Take one suffix L of the list.
        :la        Apply length to all numbers in that suffix.
           #=,     All lengths must be equal.
              Lh   Output is the first element of L.

Clojure, 63 bytes

(reduce #(if(=(quot %1 10)(quot %2 10))(max %1 %2) %1)(sort x)) 

as in:

(reduce #(if(=(quot %1 10)(quot %2 10))(max %1 %2) %1)(sort[3 7 121 11 8 2 10 9]))
=> 9

Though I'm sure there's a way to make it smaller.

JavaScript ES6, 80 77 70 bytes

a=>Math.max(...a.filter(l=>l.length==Math.min(...a.map(i=>i.length))))

I hope I am going in the right direction...

JavaScript (ES6), 62 bytes

var solution =

a=>a.map(n=>(l=`${n}`.length)>a?l>a+1|n<r?0:r=n:(a=l-1,r=n))|r

;document.write('<pre>' + `
[1] -> 1
[9] -> 9
[1729] -> 1729
[1, 1] -> 1
[34, 3] -> 3
[38, 39] -> 39
[409, 12, 13] -> 13
[11, 11, 11, 1] -> 1
[11, 11, 11, 11] -> 11
[78, 99, 620, 1] -> 1
[78, 99, 620, 10] -> 99
[78, 99, 620, 100] -> 99
[1, 5, 9, 12, 63, 102] -> 9
[3451, 29820, 2983, 1223, 1337] -> 3451
[738, 2383, 281, 938, 212, 1010] -> 938
`.split('\n').slice(1, -1).map(c =>
  c + ', result: ' + solution(eval(c.slice(0, c.indexOf('->'))))
).join('\n'))

Jelly, 7 bytes

DL,NµÞḢ

Test it at TryItOnline
Or see all test cases also at TryItOnline

How?

DL,NµÞḢ - Main link takes one argument, the list, e.g. [738, 2383, 281, 938, 212, 1010]
D       - convert to decimal, e.g. [[7,3,8],[2,3,8,3],[2,8,1],[9,3,8],[2,1,2],[1,0,1,0]]
 L      - length, e.g. [3,4,3,3,3,4]
   N    - negate, e.g [-738, -2383, -281, -938, -212, -1010]
  ,     - pair, e.g. [[3,-738],[4,-2383],[3,-281],[3,-938],[3,-212],[4,-1010]]
    µ   - make a monadic chain
     Þ  - sort the input by that monadic function, e.g [938,738,281,212,2383,1010]
          (the lists in the example are not created, but we sort over the values shown)
      Ḣ - pop and return the first element, e.g. 938

Python 2, 48 42 bytes

-6 bytes thanks to @Dennis (use min rather than sorted)

lambda l:min(l,key=lambda x:(len(`x`),-x))

All test cases are at ideone

Take the minimum of the list by (length, -value)

Pip, 11 bytes

(SNgSK-#_v)

Takes input as command-line args. Try it online!

First time using the Sort-Keyed operator! Like Python's sorted(), it takes a function that is applied to each item of the iterable and the result used as a sort key. Here's how this program works:

 SNg         List of cmdline args, sorted numerically in increasing order
    SK       Sort with key function...
      -#_    ... negative length(x), thus putting the shortest numbers at the end but not
               affecting the relative ordering among numbers with the same length
(        v)  Get the last element (index -1) and auto-print

Python 3, 56 bytes

lambda a:sorted(sorted(a),key=lambda x:-len(str(x)))[-1]

Uses a lambda in a lambda!

Python 2, 53 bytes

s=lambda a:sorted(sorted(a),key=lambda x:-len(`x`))[-1]

Same but with backticks

Jelly, 8 bytes

DL€İMị¹Ṁ

Try it online! or Verify all test cases.

Explanation

DL€İMị¹Ṁ  Input: list A
D         Convert each integer to a list of base 10 digits
 L€       Get the length of each list (number of digits of each)
   İ      Take the reciprocal of each
    M     Get the indices of the maximal values
      ¹   Get A
     ị    Select the values at those indices from A
       Ṁ  Find the maximum and return

Python 2, 58 bytes

def F(x):l={len(`i`):i for i in sorted(x)};print l[min(l)]

MATL, 14 bytes

10&YlktX<=G*X>

Try it online!

Explanation:

  &Yl           % Log
10              % Base 10
     kt         % Floor and duplicate
       X<       % Find the smallest element
         =      % Filter out elements that do not equal the smallest element
          G     % Push the input again
           *    % Multiply (this sets numbers that do not have the fewest digits to 0)
            X>  % And take the maximum

Ruby, 34 bytes

->a{a.max_by{|n|[-n.to_s.size,n]}}

See it on eval.in: https://eval.in/643153