co.combinatorics - Balancing problem

There was a problem in an Olympiad selection test, which went as follows: Consider the set ${1,2,dots,3n }$ and partition it into three sets A, B and C of size n each. Then, show that there exist x, y and z, one in each of the three sets, such that x + y = z.

This has a tricky-to-get but otherwise straightforward solution, that starts by assuming 1 to be in A, finding the smallest k not in A, assuming that to be in B, and then arguing that no two consecutive elements can be present in C (for that would give an infinite descent). Finally, cardinality considerations solve the problem.

I managed to prove a corresponding statement for 4n, namely: for ${ 1,2,3, dots, 4n }$, partitioned into four sets of size n each, there exist x, y, z, and w, one in each set, such that x + y = z + w.

The question here is whether analogues of this hold for all m, with $m ge 3$ and $n ge 2$. In other words, if ${ 1,2, dots, mn }$ is divided into $m$ sets of size $n$ each, can we always make a choice of one element in each set such that the sum of floor $m/2$ of the elements equals the sum of the remaining ceiling $m/2$ elements ($(m-1)/2$ and $(m + 1)/2$ for $m$ odd, $m/2$ each for $m$ even). Note we need $n ge 2$ due to parity considerations when $m$ is congruent to $1$ or $2$ modulo $4$.