if you're looking for numbers up to 111,111,111, then the code is

numbersWithCrosssum[n_Integer] :=
 Select[Range[111111111], Total@IntegerDigits[#] == n &]

numbersWithCrosssum[36]

or

Table[If[Total@IntegerDigits[i] == 36, i, {}], {i, 1, 111111111}] // Flatten

While this is the most idiomatic way of solving your problem, it is not the most efficient (re memory, time) because any implementation involving Select[ ] or Cases[ ] is horribly inefficient, but i will refuse to present a more efficient implementation. After all this solution does work and is the most idiomatic. This should be enough, bye.

I guess in this context the function IntegerPartitions should at least be mentioned. It could be used here like so:

numberList[n_] := Select[IntegerPartitions[n], Max[#] < 10 &]

and from this calculate the numbers, e.g.:

nl = numberList[12];
FromDigits /@ Flatten[Permutations /@ nl, 1]

I cannot imagine to calculate all those numbers for non small n, because the length of this list goes approximately like 0.50398 E^(0.692161 n), which e.g. for n=36 gives 3.34258*10^10 different numbers. Or am I misunderstanding the question?

Hi Daniel! I want to find numbers (quickly) which have always 36 as cross sum, like 9999 or 131923179. I'm looking for a method that generates me quickly such numbers, or a method which generates me directly a list of numbers which have always 36 as cross sum, like the number 9999 or 131923179. I hope you can help me. Best regards.