I'm not sure of the exact performance implications, but using CoprimeQ with just one pass might be better than creating all of the fractions and then selecting based on the length of the denominator. Using a helper function,

AllFractionsWithDenom[d_] := Select[Range@d, CoprimeQ[#, d] &]/d

and then timing this

Flatten[AllFractionsWithDenom /@ Range[1000, 9999]]

takes about 55 seconds on my machine.

On the other hand, timing this

Union[Divide @@@ Flatten[Table[{m, k}, {k, Range[10^4 - 1]}, {m, k - 1}], 1]]

(which doesn't include the next step to filter down to just 4 digit denominators) takes 259 seconds on my machine.

StackOverflow, at the link above, has already provided a few very fast implementations; here is another one. It uses the CoprimeQ trick from @Eric Rimbey and the fact that if m/n is an x-coordinate of a rational point on a uni circle, then n^2-m^2 is an exact square. We can then hash all the exact squares once to speed up the calculation considerably.

ClearAll[RationalUni, DenominatorLength, PythagoreanTuples]

DenominatorLength[x_] := 
 If[IntegerQ[x], 0, Ceiling[Log10[Denominator[x]]]]

PythagoreanTuples[n_, squares_] :=
 {#/n, Sqrt[1 - (#/n)^2]} & /@
  Select[Range[1, Ceiling[Sqrt[n^2/2]]], 
   CoprimeQ[n, #] && squares[n^2 - #^2] &]

RationalUni[n_] := Module[{squares},
  squares[_] = False;
  Scan[(squares[#^2] = True) &, Range[10^n]];
  Union@Flatten@Select[
     Flatten[
      Parallelize[
       PythagoreanTuples[#, squares] & /@ 
        Range[10^(n - 1), 10^n - 1]], 1],
     DenominatorLength[Last[#]] == n &]]

Let's test it:

In[220]:= Sort@ratUni[3] == Sort@RationalUni[3]
Out[220]= True

In[221]:= ratUni[4] // Length // AbsoluteTiming
Out[221]= {970.987, 2870}

In[219]:= RationalUni[4] // Length // AbsoluteTiming
Out[219]= {7.80111, 2870}

In[213]:= RationalUni[5] // Length // AbsoluteTiming
Out[213]= {739.865, 28652}

We managed to speed up the calculation for n=4 by over 100 times, measuring on the same CPU, and calculate RationalUni[5] in just over 12 minutes, which is still faster than the original solution for n=4.

macOS Big Sur 11.7.2 / Wolfram Desktop 13.2 /
3.6 GHz 8-Core Intel Core i9

@Henrik - wonderful idea - thank you very much!

...and @Victor -- thanks a lot for computing RationalUni[5] -- it was unattainable for me. I wonder what the next few points are for $n=6,7,8,...etc.$ but we'd probably need a cluster for that. Time seems rapidly increasing.

E.g.

In[5]:= ParallelTable[{a, b}/c /. # & /@ 
       Solve[a^2 + b^2 == c^2 && a <= b, PositiveIntegers], {c, 5, 
       100000, 4}] // Flatten[#, 1] & // SortBy[Denominator@*First] //
    DeleteDuplicates // Length // AbsoluteTiming
Out[5]= {7.43017, 15919}

Which matches the total of the first five elements of your sequence below:

In[4]:= Total@{1, 15, 142, 1435, 14326}
Out[4]= 15919