Prove that you can choose $ 2 ^ {k-1} $ numbers from $2^k -1 $ integers so that the sum of them is divisible by $2^{k-1}$

Question

Prove that you can choose $ 2 ^ {k-1} $ numbers from $2^k -1 $ integers so that the sum of them is divisible by $2^{k-1}$ .

The work I have done :

I wanted to use strong induction , I presumed the theorem is correct for all numbers less than $2^k -1 $ in the form of $2^n -1 $ from there I concluded you can choose $2^{k-1} $ numbers from our $2^k -1 $ numbers and you can write them in the form of $ 2 ^ {k-1} . m $ , the remaining numbers will be $ 2 ^ {k} $- $ 2 ^ {k-1} : { 2 ^ {k-1} }$ we can write that in the form of $ 2 ^ {k-1} $ too , but at this point I realized I was way off the solution and appreciate any other ideas (even not involving induction)

Note another duplicate post is West German Mathematics Olympiad 1981. Also, an almost duplicate post is Prove for any number n, it is possible to select $X = 2^n$ numbers from $2^{n+1}$ numbers s.t. the sum of X is divisible by $2^n$ (with this one using a set size one larger than yours). — John Omielan, Jun 22 '21 at 07:37
Do you have any advice for searching these solutions up ? I cannot find the problems related to my own problem without not posting and hearing other people 's links . — s1234s1234d95829, Jun 22 '21 at 07:54
Thank you for your interest & concern re: searching for existing appropriate posts. My advice is to first read How to search on this site?, & possibly also a few other posts linked there in the comments & answers. I don't have much to add except that the built-in search uses "and" between terms, so try to keep the length & number of terms reasonably minimal. Finally, FYI, I was able to come up with the duplicate of your question, & the other $2$ posts, so quickly as they're in my previous answer. — John Omielan, Jun 22 '21 at 20:20

Prove that you can choose $ 2 ^ {k-1} $ numbers from $2^k -1 $ integers so that the sum of them is divisible by $2^{k-1}$

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