1967 IMO Problems/Problem 5
Take . Suppose that are all equal and greater than . Then for sufficiently large , we can ensure that for , and hence the sum of for all is less than . Hence must be even with half of positive and half negative.
If that does not exhaust the , then in a similar way there must be an even number of with the next largest value of , with half positive and half negative, and so on. Thus we find that for all odd .