TI-BASIC (TI-84 Plus CE Python), 74 67 65 bytes

While 1
Ans+1
If not(fPart(round(AnsΣ(int(Ans/K)-int((Ans-1)/K),K,1,1+Ans)/Σ(Knot(fPart(Ans/K)),K,1,1+Ans),9
Disp Ans
End

make sure Ans is set to 0 before running.

jBasher2, 592 bytes

create n with type number
while 1 == 1
add n by 1
set that to n
create k with type number
set 1 to k
create f with type number
create o with type number
while k <= n
create g with type number
subtract n by 1
divide that by k
parse that as int
set that to g
divide n by k
parse that as int
subtract that by g
add f by that
set that to f
divide n by k
modulo that by 1
if that == 0
add k by o
set that to o
endif
add 1 by k
set that to k
endwhile
divide 1 by 10
add o by that
parse that as int
set that to o
divide f by o
multiply that by n
modulo that by 1
if that == 0
output n
endif
endwhile

awesome translation of my YASEPL answer

APL(NARS), 80 chars

r←F w;x;m;k
r←⍬⋄x←0x
→0×⍳0≥w
x+←1⋄→3×⍳∼1=÷1∨(≢m)÷+/÷m←k/⍨0=x∣⍨k←⍳x
r,←x⋄w-←1⋄→2

// +/ 12 9 8 38 13=80

F 1-indexed, and it is slow it seems has at last O(n^5) if goes well. F use rationals.
÷1∨k finds the denominator of k rational.

Test:

  F 5
1 6 28 140 270 
  F 10
1 6 28 140 270 496 672 1638 2970 6200 
  F 1
1 

YASEPL, 81 bytes

=n!+=k$=f=o`1=g$n-/k(=l$n/k(-g!f+l=u$n/k%1+[2!o+k`2!k+}4,n!o+.9(!f/o*n%1+[3>n`3?3

this prints them infinitely

explanation

=n                        main increment
                          (after this is where the loop starts)
  !+                        add 1 to n
  =k$                       K variable for summation
  =f                        F is total amount of divisors of N
  =o                        O is sum of divisors of N
  `1                        start sum
                              ( calculating number of divisors )
    =g$n-/k(=l$n/k(-g         l = floor(n/k) - floor(n-1/k)
    !f+l                      add to sum
                              ( calculating sum of divisors )
    =u$n/k%1+[2               if (n/k) = 0...
      !o+k                    add K to O
    `2                      end if
    !k+                     increment K
  }4,n                      if K <= N, go back to `1
  !o+.9(                    ceil O
  !f/o*n%1+[3               if (f/o)*n = 0...
    >n                        print n
  `3                        endif
?3                        go back to the third char of the program (where !+ is)     

this assumes that \$ n \$ is a harmonic divisor number if the following is true:

$$ 0=(\frac{\sum_{k=1}^{n}(\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k} \right\rfloor)}{\left\lceil\sum_{i=1}^{n}(0^{\frac{n}{i}\mod1}i)\right\rceil}\times n)\mod1 $$ I'm not the best with writing math equations so take it with a grain of salt.

this thing gets really slow.

Nibbles, 8.5 bytes

|,~~%*,;|,$~%@$@+

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Pyt, 22 bytes

1`ĐðĐĐΠ⇹/Ʃ⇹Π|?ŕĐƥ:ŕ;⁺ł

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Prints Ore numbers forever.

1                              Push 1
 `                   ł         Do... while top of stack is truthy
  Đ                            Đuplicate
   ð                           Get list of ðivisors
    ĐĐ                         Đuplicate twice
      Π                        Get product
       ⇹                       Swap top two elements on stack
        /                      Divide
         Ʃ                     Ʃum
          ⇹                    Swap top two elements
           Π|?                 Does the sum divide the product cleanly?
              ŕĐƥ              If so, ŕemove the boolean, Đuplicate, and ƥrint
                 :ŕ            Otherwise, ŕemove the boolean
                   ;⁺          Either way, increment

Vyxal 2.6.1, 5 bytes

≬KṁḊȯ

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This doesn't work in 2.4 because 2.6 uses sympy rationals to store non-integers. Essentially a port of hyper's hypothetical Jelly answer.

Explained

≬KṁḊȯ
≬     # The next three elements as a function, taking single argument n:
 K    #    divisors of n
  ṁ   #    the average of that
   Ḋ  #    does that divide n?
    ȯ # First input numbers that satisfy the above function.

Python 3, 94 bytes

v=0
while 1:p=q=1;v+=1;-~len([(p:=p*d+q,q:=q*d)for d in range(2,v+1)if v%d<1])*q%p or print(v)

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Outputs indefinitely. This is otherwise like (my edit to) @Jitse's answer, but computes the sum of the reciprocals p/q exactly as a pair of (big)ints (p,q).

Scala, 111 bytes

Stream.iterate(1:BigInt)(_+1)filter{n=>val d=n to(1,-1)filter(n%_<1)
val p=d.product
p*d.size%d.map(p./).sum<1}

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Returns an infinite Stream.

Scala, 92 bytes

Stream from 1 filter{n=>val d=1 to n filter(n%_<1)
val p=d.product
p*d.size%(0/:d)(_+p/_)<1}

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This one uses normal Ints, evading some of the boilerplate above, but it only generates the first three elements correctly due to integer overflows.

Zephyr, 151 bytes

set n to 1
while 1=1
set s to 0
set c to 0
for d from 1to n
if(n mod d)=0
set s to(/d)+s
inc c
end if
next
if((c/s)mod 1)=0
print n
end if
inc n
repeat

Try it online! Uses the output-infinitely strategy; you'll need to kill the program before 60 seconds in order to see any output.

Ungolfed

# Start from 1
set num to 1
# Loop forever
while true
  # Calculate the sum of the reciprocals of the divisors
  # and also the total number of divisors
  set reciprocalSum to 0
  set divisorCount to 0
  for divisor from 1 to num
    if (num mod divisor) = 0
      set reciprocalSum to reciprocalSum + (/ divisor)
      inc divisorCount
    end if
  next
  # Print the number if the divisor count divided by the
  # divisor-reciprocal sum is an integer
  if ((divisorCount / reciprocalSum) mod 1) = 0
    print num
  end if
  # Go to the next number
  inc num
repeat

Wolfram Language (Mathematica), 42 40 bytes

-2 bytes thanks to att!

Do[Mean@Divisors@n∣n&&Print@n,{n,∞}]

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Stax, 11 bytes

¡♪♫ö╪ü♣↕¥Vv

Run and debug it

Runs an infinite loop with no input.

Husk, 9 bytes

fo§¦ṁ\LḊN

Try it online! (header outputs the first few elements to avoid timing-out)

         N  # from the infinite list of integers
fo          # output those for which
     ṁ\     # the sum of the reciprocals of their divisors
  §¦        # exactly divides
       LḊ   # the length (number) of their divisors

Core Maude, 248 bytes

mod H is pr LIST{Rat}. ops o h : Rat ~> Rat . var A B C D : Rat . eq o(A)= o(2
A). eq o(s A 0)= A . eq o(A s B)= o(s A(B + ceiling(frac(h(A A 0 0))))). eq h(A
s B C D)= h(A B(0 ^(A rem s B)/ s B + C)(0 ^(A rem s B)+ D)). eq h(A 0 C D)=
D / C . endm

The result is obtained by reducing the o function with the zero-indexed input \$n\$.

Example Session

Maude> red o(0) .  --- 1
result NzNat: 1
Maude> red o(1) .  --- 6
result NzNat: 6
Maude> red o(2) .  --- 28
result NzNat: 28
Maude> red o(3) .  --- 140
result NzNat: 140
Maude> red o(4) .  --- 270
result NzNat: 270
Maude> red o(5) .  --- 496
result NzNat: 496
Maude> red o(6) .  --- 672
result NzNat: 672
Maude> red o(7) .  --- 1638
result NzNat: 1638
Maude> red o(8) .  --- 2970
result NzNat: 2970
Maude> red o(9) .  --- 6200
result NzNat: 6200
Maude> red o(10) .  --- 8128
result NzNat: 8128
Maude> red o(11) .  --- 8190
result NzNat: 8190

Ungolfed

mod H is
    pr LIST{Rat} .

    ops o h : Rat ~> Rat .

    var A B C D : Rat .

    eq o(A) = o(2 A) .
    eq o(s A 0) = A .
    eq o(A s B) = o(s A (B + ceiling(frac(h(A A 0 0))))) .

    eq h(A s B C D) = h(A B (0 ^ (A rem s B) / s B + C) (0 ^ (A rem s B) + D)) .
    eq h(A 0 C D) = D / C .
endm

Maude has built-in support for rational arithmetic, so we just compute the harmonic mean of the divisors with h. Then, ceiling(frac(h(...))) will be 0 if h(...) is a natural number or 1 otherwise. Also, note that in Maude 0 ^ 0 == 1 and 0 ^ X = 0 for X =/= 1.

Jelly, 9 bytes

1Æd×ọÆsƲ#

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More boring than the other answer:

             Implicit input: an integer z.
1      Ʋ#    Count up from 1, finding z numbers for which...
 Æd×           divisor_count(n) × n
    ọ          is divisible by
     Æs        divisor_sum(n).

Ruby, 71 56 bytes

1.step{|n|k=0;(1..n).count{|x|n%x<1&&k+=1r/x}%k>0||p(n)}

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• Saved 15 thanks to @G B lots of golfs

Outputs the sequence indefinitely.

JavaScript (V8), 63 bytes

Prints the sequence forever.

{for(n=0;;s%t||print(n))for(k=++n,t=s=0;k;)n%k--||(s+=n,t-=~k)}

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Python 3, 79 bytes

n=0
while 1:n+=1;a=[i for i in range(1,n+1)if n%i<1];n*len(a)%sum(a)or print(n)

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Outputs indefinitely.

R, 55 52 bytes

-3 bytes thanks to @Dominic van Essen.

while(F<-F+1)(1/mean(1/(y=1:F)[!F%%y]))%%1||print(F)

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Prints Ore numbers infinitely.

MathGolf, 13 bytes

î∙─‼Σ£î*\÷╛p∟

Outputs indefinitely.

Try it online. (You do have to manually cancel it during runtime to see output apparently, before it times out after 60 seconds..)

Explanation:

            ∟  # Do-while true without popping:
î              #  Push the 1-based loop-index
 ∙             #  Triplicate it
  ─            #  Pop the top, and get a list of its divisors
   ‼           #  Apply the following two commands separately:
    Σ          #   Sum the divisors-list
     £         #   Get the length of the divisors-list
      î*       #  Multiply the length by the 1-based loop-index
        \      #  Swap the top two values on the stack
         ÷     #  Check that the length*î is divisible by the sum
          ╛    #  If this is truthy:
           p   #   Pop the remaining copy of the index, and print it

Raku, 46 bytes

grep {{@_%%sum 1 X/@_}(grep $_%%*,1..$_)},^∞

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This is a lazy infinite sequence of the harmonic divisor numbers.

Factor + lists.lazy math.primes.factors math.unicode, ⁶⁹ 65 bytes

[ 1 lfrom [ divisors [ length 1 ] keep n/v Σ mod 0 = ] lfilter ]

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It's a quotation that returns an infinite lazy list of the harmonic divisor numbers.

Explanation

• 1 lfrom an infinite lazy list of natural numbers
• [ ... ] lfilter select numbers for which the quotation returns true
• divisors get the divisors of a number (e.g. 6 divisors -> { 1 2 3 6 })
• [ length 1 ] keep (e.g. { 1 2 3 6 } [ length 1 ] keep -> 4 1 { 1 2 3 6 })
• n/v divide number by vector (e.g. 1 { 1 2 3 6 } n/v -> { 1 1/2 1/3 1/6 })
• Σ take the sum
• mod 0 = is it a divisor?

Pari/GP, 42 bytes

for(n=1,oo,numdiv(n)*n%sigma(n)||print(n))

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Jelly, 9 bytes

ÆDpWSḍ/µ#

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This is horribly scuffed because I couldn't figure out how to get it working with precision. I had the same idea as ovs it turns out, but ÆDÆmḍµ# fails due to precision issues.

I honestly hate how this is written.

ÆDpWSḍ/µ#    Main Link
       µ#    nfind; return first n values satisfying:
ÆD           divisors of n
  p          cartesian product with
   W         [n] (returns [[a, n], [b, n], ...])
    S        sum (returns [divisor sum, divisor count * n])
     ḍ/      reduce by divisibility check

Charcoal, 37 bytes

Ｎθ≔⁰ηＷθ«≦⊕η≔Φ⊕η∧κ¬﹪ηκζ¿¬﹪×ηＬζΣζ≦⊖θ»Ｉη

Try it online! Link is to verbose version of code. Outputs the 1-indexed nᵗʰ Ore number. Explanation:

Ｎθ

Input n.

≔⁰η

Start looking for Ore numbers greater than zero.

Ｗθ«

Repeat until the nᵗʰ number has been found.

≦⊕η

Try the next integer.

≔Φ⊕η∧κ¬﹪ηκζ

Get its factors.

¿¬﹪×ηＬζΣζ

If the harmonic mean is an integer, then...

≦⊖θ

Decrement the count of remaining Ore numbers to find.

»Ｉη

Print the found Ore number.