The puzzle, originally, was* if you have N numbers, to show that some sum of some of them has total to a multiple of N.
*

I think I might have changed this, in midstream, to

*If you have N numbers, show that some sum of them has to total to a multiple of N-1.*

This is even easier, and in fact generalizes. For any M less than or equal to N, the pigeonhole principle gives us the proof, though we need a slight twist when M=N (i.e. for the original problem):

For the N numbers a,b,c,… we first list the sums a, a+b, a+b+c, etc.

Now then, if we consider M less than N, these sums can only total to 0, 1, 2, …, (M-1) (mod M) As there are N pigeons (namely a, a+b, etc) and M holes (0,1,.., M-1) at least two pigeons have to go in the same hole.

Suppose a+b+c and a+b+c+d+e are the same mod M. Then their difference, d+e, is 0 mod M, and so is a multiple of M.

If M actually equals N, as in the original problem the argument doesn’t necessarily work and we need a slight variation on the proof — the proof above will work, unless the N pigeons exactly fill up the N holes 0,1, 2, … (N-1). But then some pigeon has to fill up hole 0; that is, some sum has to total 0 mod N, and again we are done.

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