Denote the sum of all the elements in by .
then for all proper subsets of .
So for any proper subset of there were admissible values of .
Now apply the Pigeon Hole Principle (a.k.a the dirchilet's box principle).
we had proper subsets and had only possible values of . So according to the Pigeon Hole Principle there are two proper subsets and of such that . If we are done. Else if then consider the proper subsets and that is delete the common elements from both and .