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
.