Japt -h, 21 9 7 bytes

Ever have one of those challenges where you completely forget how to golf?!
Much better :D

à có ñx

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à có ñx     :Implicit input of array
à           :Combinations
  c         :Flat map
   ó        :  Uninterleave
     ñ      :Sort by
      x     :  Sum
            :Implicit out put of last sub-array

JavaScript (Node.js), 91 bytes

g=([p,q,...x],n=N=-1/0,...s)=>1/p?g(x,~~n+p,...s,p)&&g([q,...x],n,...s):N>n|s<g?S:(N=n,S=s)

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JavaScript (Node.js), 95 94 bytes

a=>(N=g=([p,q,...x],n,...s)=>1/p?g(x,n+p,...s,p)&&g([q,...x],n,...s):N>n|s<g?S:(N=n,S=s))(a,0)

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Brachylog, 11 bytes

{⊇ġ₂hᵐ}ᶠ+ᵒt

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Back-ported from ais523's Jelly solution.

{     }ᶠ       Find every way to:
 ⊇             take a subsequence of the input,
  ġ₂           split it into pairs of adjacent elements (last element can be unpaired),
    hᵐ         and keep only the first element of each pair.
        +ᵒt    Yield the one with the greatest sum.

Wolfram Language (Mathematica), 70 63 58 43 42 40 bytes

-#&@@Subsets[-#][[2;;,;;;;2]]~SortBy~Tr&

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Port of ais523's Jelly solution.

I don't think I've ever seen this many ;,s in a row before.

Jelly, 8 bytes

ŒPm€2SÞṪ

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This is a function submission (it's almost usable as a full program, which is what the TIO link does, but the output format is a little weird; when used as a function, though, it just outputs the list of elements from the subsequence).

Explanation

I found a much terser algorithm (loosely inspired by my previous Brachylog answer), saving 6 bytes over the Jelly and Brachylog answers and 3 bytes over the current leading answer.

ŒPm€2SÞṪ
ŒP         On the set of all subsequences of {the input},
  m 2        remove every second element of
   €         each subsequence,
      ÞṪ     then {return} the resulting subsequence with the maximal
     S       sum

The basic idea is that we don't look for the subsequence we want directly; rather, we find a subsequence with an additional element inserted in between each of the elements we want. This is always possible if the elements of the subsequence we want are not adjacent in the original list (because there'll always be at least one in between to insert), and never possible if the elements of the subsequence we want are adjacent in the original list. So all we have to do is look at all the subsequences of the input, then take alternating elements from them, to get a list of all the non-adjacent-item subsequences of the input.

This algorithm will return many of the non-adjacent-item subsequences more than once, but that doesn't matter, because we're just looking for a maximal element, and repeating elements of the set we're taking a maximum of has no influence on the maximum.

T-SQL, 122 119 118 bytes

Input is a table variable.

This query picks all elements from the table variable, combining these with all non-adjacent elements with higher position values and show the text generated for the highest sum of these values.

WITH C(y,j,v)as(SELECT*,x*1FROM @
UNION ALL
SELECT y+','+x,i,v+x
FROM @ JOIN C ON~-i>j)SELECT
TOP 1y FROM C ORDER BY-v

Try it online ungolfed

J, 47 bytes

>@{:@(<@#~/:+/@#~)1(-#~0=1 1+/@E."1-)[:i.&.#.=~

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-7 bytes thanks to FrownyFrog

original

J, 54 bytes

[:(>@{:@/:+/&>)]<@#~"1[:(#~0=1 1+/@E."1])[:#:@}.@i.2^#

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R, 108 107 bytes

function(v,M=-Inf){for(j in J<-seq(a=v))for(i in combn(J,j,,F))if(all(diff(i)>1)&sum(v[i])>sum(M))M=v[i]
M}

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-1 thanks to @Giuseppe

R, 136 125 bytes

function(l,G=unlist(Map(combn,list(y<-seq(a=l)),y,c(function(x)'if'(all(diff(x)>1),l[x],-Inf)),F),F))G[which.max(Map(sum,G))]

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-6 bytes thanks to digEmAll, who incidentally also outgolfed me.

Returns the shortest subsequence as a list, breaking ties on lexicographically first by indices.

Brute-force generates all index subsequences, then Filters for those that are non-adjacent, i.e., where all(diff(x)>1). Then subsets [ into l using these indices, selecting [[ the first one where the sum is the max (which.max).

I'm pretty sure this is the first R answer I've ever written that uses Filter! sad, Filter is ungolfy, no wonder I've never used it...

Python 2, 63 70 65 bytes

f=lambda a:a and max([a[:1],a[:1]+f(a[2:]),f(a[1:])],key=sum)or a

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5 bytes thx to ArBo

Haskell, 60 bytes

snd.([]%)
r%(h:t)=max(r%t)$(r++[h])%drop 1t
r%_=(sum r<$r,r)

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The helper function % recursively branches on choosing whether the include the first element and drop the second, or to skip the first element. It takes the maximum of all outcomes, which are tuples whose first element is the sum, and whose second element is the corresponding list which is extracted for the output.

To handle the rule that the empty list is disallowed even if it would have the smallest trick, we do a cute trick of writing sum r<$r rather than sum r.This makes a list whose elements all are sum r and whose length is that of r. That way, when we choose the maximum, we prioritize any list over an empty r, but otherwise comparisons depend on the first element which is sum r .

Japt -h, 22 bytes

Êo à k_mÄ øZÃm!gU
fÊñx

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Gaia, 24 bytes

e:w;ċz⟨ọ1>¦ẏ⟩⁇‼⁇E‡ev2%Σ⌠

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Ugh, E‡ does some weird stuff...according to the documentation, it should do something like "given length i set of lists X and length j set of indices Y, return X[i][Y[j]]", but instead returns [X[i][Y[j]] X[i][Y[-j]] where negative indexing represents the complement, so we have to do ev2% to extract only the ones we want.

e				| eval as a list l
 :				| dup
  w				| wrap as a list
   ;				| push l again
    ċ				| push [1..len(l)]
     z				| push all subsets of [1..len(l)] -- index powerset.
      ⟨      ⟩⁇			| filter this for:
       ọ			| deltas
        1>¦			| are greater than 1
           ẏ			| all (all deltas greater than 1)
	       ‼⁇		| filter for non-empty lists
		 E‡		| table extract elements. Given l and index set i, this pushes
				| [l[i] l[setdiff(1..l,i)]] for some reason
		   ev2%		| get the l[i] only by unlisting, reversing, and taking every other element
		       Σ⌠	| Get the one with the maximum sum

Haskell, 81 80 bytes

snd.maximum.map((,)=<<sum).tail.f
f(a:b:c)=f(b:c)++map(a:)(f c)
f a=[]:map(:[])a

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f builds all valid subsequences by either skipping the next element (f(b:c)) or using it and skipping the next (map(a:)(f c)) and recursively work on the rest. For the result, build all subsequences (f), drop the empty subsequence (which occurs first in the list: tail), make pairs (<sum>,<subsequence>) (map((,)=<<sum)), find the maximum (pairs are compared in lexicographical order) -> maximum) and drop the sum (snd).

Edit: -1 byte thanks to @Lynn.

JavaScript (ES6),  138 132 130 129  126 bytes

Outputs key-value pairs.

a=>a.reduce((a,x,i)=>[...a,...a.map(y=>[[x,i],...y])],[[]]).map(m=a=>a.some(s=p=([v,i])=>p-(s=~~s+v,p=i)<2)|s<m||(r=a,m=s))&&r

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Step 1

We first compute the powerset of the input with \$[value, index]\$ pairs.

a.reduce((a, x, i) => // for each value x at position i:
  [                   //   update a[] to a new array consisting of:
    ...a,             //     all previous entries
    ...a.map(y =>     //     for each value y in a[]:
      [[x, i], ...y]  //       append [x, i], followed by all original entries
    )                 //     end of map()
  ],                  //   end of new array
  [[]]                //   start with a = [[]]
)                     // end of reduce()

Step 2

We then look for the maximum sum \$m\$ among these sets, discarding sets with at least two adjacent elements. The best set is stored in \$r\$.

.map(m =              // initialize m to a non-numeric value
  a =>                // for each entry a[] in the powerset:
  a.some(s = p =      //   initialize s and p to non numeric values
    ([v, i]) =>       //   for each value v and each index i in a[]:
    p - (             //     compute p - i
      s = ~~s + v,    //     add v to s
      p = i           //     update p to i
    ) < 2             //     if p - i is less than 2, yield true
  ) |                 //   end of some()
  s < m ||            //   unless some() was truthy or s is less than m,
  (r = a, m = s)      //   save a[] in r[] and update m to s
) && r                // end of map(); return r[]

Charcoal, 46 bytes

≔⟦υ⟧ηＦθ«≔υζ≔Ｅη⁺κ⟦ι⟧υ≔⁺ζηη»≔Φ⁺υηιη≔ＥηΣιζＩ§η⌕ζ⌈ζ

Try it online! Link is to verbose version of code. Explanation:

≔⟦υ⟧η

The variable u is predefined with an empty list. This is put in a list which is assigned to h. These variables act as accumulators. u contains the sublists that include the latest element of the input q while h contains the sublists that do not (and therefore are suitable for appending the next element of the input).

Ｆθ«

Loop over the elements of the input.

≔υζ

Save the list of sublists that contain the previous element.

≔Ｅη⁺κ⟦ι⟧υ

Take all of the sublists that do not contain the previous element, append the current element, and save the result as the list of sublists that contain the current element. (I don't use Push here as I need to clone the list.)

≔⁺ζηη»

Concatenate both previous sublists into the new list of sublists that do not contain the current element.

≔Φ⁺υηιη

Concatenate the sublists one last time and remove the original empty list (which Charcoal can't sum anyway).

≔ＥηΣιζ

Compute the sums of all of the sublists.

Ｉ§η⌕ζ⌈ζ

Find an index of the greatest sum and output the corresponding sublist.

Jelly, 16 14 bytes

JŒPḊf’$ÐḟịµSÞṪ

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Thanks to @EriktheOutgolfer for saving 2 bytes!

Husk, 11 bytes

►Σ†!¹mü≈tṖŀ

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Explanation

►Σ†!¹mü≈tṖŀ  Implicit input, say L=[4,5,3,4].
          ŀ  Indices: [1,2,3,4]
         Ṗ   Powerset: [[],[1],[2],[1,2],..,[1,2,3,4]]
        t    Tail (remove the empty list): [[1],[2],[1,2],..,[1,2,3,4]]
     m       For each,
      ü      de-duplicate by
       ≈     differing by at most 1.
             For example, [1,2,4] becomes [1,4].
  †          Deep map
   !¹        indexing into L: [[4],[5],[4],..,[5,4],[4,3]]
►            Maximum by
 Σ           sum: [5,4]

Brachylog (v2), 14 bytes

{~ba~c∋₁ᵐ}ᶠ+ᵒt

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Function submission; input from the left, output from the right, as usual. Very slow; a five-element list is probably the maximum for testing on TIO.

{~ba~c∋₁ᵐ}ᶠ+ᵒt
 ~b              Prepend an arbitrary element to the input
   a             Take a prefix or suffix of the resulting list
    ~c           Ordered partition into contiguous sublists
      ∋₁         Take the second element
        ᵐ          of each sublist
{        }ᶠ      Find all possible ways to do this
           +ᵒ    Sort by sum
             t   Take the greatest

The results we get from prefixes aren't incorrect, but also aren't interesting; all possible results are generated via taking a suffix (which is possibly the list itself, but cannot be empty), but "suffix" is more verbose in Brachylog than "prefix or suffix", so I went with the version that's terser (and less efficient but still correct). The basic idea is that for each element we want in the output list, the partition into contiguous sublists needs to place that element and the element before into the same sublist (because the element is the second element of the sublist), so two consecutive elements can't appear in the result. On the other hand, it's fairly clear that any list without two consecutive elements can appear in the result. So once we have all possible candidate lists, we can just take the sums of all of them and see which one is largest.

Haskell, 300 168 bytes

import Data.List
h[]=1>2
h(x:y)=fst$foldl(\a c->((fst a)&&(c-snd a>1),c))(1<2,x)y
z=snd.last.sortOn fst.map((,)=<<sum.snd).filter(h.fst).map unzip.subsequences.zip[0..]

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-132 bytes thanks to all the feedback from @nimi :)

Original

Ungolfed (original)

import Data.List
import Data.Function

f :: [Int] -> [(Int, Int)] -- attach indices for later use
f [] = []
f xs = zip xs [0..length xs]

g :: [[(Int, Int)]] -> [([Int], [Int])] -- rearrange into list of tuples
g [] = []
g (x:xs) = (map fst x, map snd x) : g xs

h :: [Int] -> Bool -- predicate that checks if the indices are at least 2 apart from each other
h [] = False
h (x:xs) = fst $ foldl (\acc curr -> ((fst acc) && (curr - snd acc > 1), curr)) (True, x) xs
j :: [([Int], [Int])] -> [([Int], [Int])] -- remove sets that don't satisfy the condition
j xs = filter (\(elements, indices) -> h indices) xs

k :: [([Int], [Int])] -> [(Int, ([Int], [Int]))] -- calculate some of elements
k xs = map (\(elements, indices) -> (foldl1 (+) elements, (elements, indices))) xs

l :: [(Int, ([Int], [Int]))] -> ([Int], [Int]) -- grab max
l xs = snd $ last $ sortBy (compare `on` fst) xs

z -- put things together
```

05AB1E, 14 bytes

Saved 1 byte thanks to Kevin Cruijssen

ā<æʒĆ¥≠W}èΣO}θ

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Explanation

ā<               # push [0 ... len(input)-1]
  æ              # compute powerset
   ʒ    }        # filter, keep lists where:
      ≠W         # no element is 1 in the
     ¥           # deltas
    Ć            # of the list with the head appended
         è       # index into the input with each
          ΣO}    # sort by sum
             θ   # take the last element

Pyth, 19 bytes

esDm@LQdtf!q#1.+TyU

Try it online here, or verify all the test cases at once here.

esDm@LQdtf!q#1.+TyUQ   Implicit: Q=eval(input())
                       Trailing Q inferred
                  UQ   Generate range [0-len(Q))
                 y     Take the powerset of the above
         f             Filter keep elements of the above, as T, using:
              .+T        Take differences of consecutive elements of T
           q#1           Keep those differences equal to 1
          !              Logical NOT - empty lists evaluate to true, populated ones to false
                       Result of the filter is those sets without consecutive numbers
        t              Drop the first element (empty set)
   m                   Map the remaining sets, as d, using:
     L d                 For each element of d...
    @ Q                  ... get the element in Q with that index
 sD                    Order the sets by their sum
e                      Take the last element, implicit print

Wolfram Language (Mathematica), 144 bytes

If[Max[a=#]>(d=Max[m=Tr/@(g=a[[#]]&/@Select[Subsets[Range[x=Length@#],{2,x}]&@#,FreeQ[Differences@#,1]&]&@a)]),{Max@a},g[[#]]&@@@Position[m,d]]&

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