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 $4$.



Balancing problem
[This](http://www.emis.de/journals/INTEGERS/papers/a9int2003/a9int2003.pdf) article is a nice survey of "Rainbow Ramsey theory". In this jargon what you are trying to prove is that the vector $(1,1,\dots,1,-1,-1,\dots,-1)$ is rainbow partition $m$-regular.


The case of $(1,1,-1,-1)$ being rainbow partition 4-regular, was proved in "Rainbow solutions for the sidon equation x+y=z+w", J. Fox, M. Mahdian, and R. Radoicic. They actually proved that as long as each of the four parts of $[n]$ has at least $(n+1)/6$ members then one can always find rainbow solutions to $x+y=z+w$ (i.e. each variable coming from a different partition.)


Though these results were mostly inspired from their monochromatic version (the 3 variable case dates back to Schur, and then R.Rado classified all linear equations that are partition regular), the analogy hasn't proven very faithful. The rainbow Hales-Jewett theorem is false, and so is the rainbow Van der Waerden theorem! 


Another thing worth mentioning is that if we color $\mathbb{Z}/p\mathbb{Z}$ in $k$ colors with each color having at least $k$ elements, then the equation $\sum\_{i=1}^k a\_i x\_i\equiv b\pmod{p}$ always has a rainbow solution given that not all $a\_i$ are the same. A proof is in [this](http://www.dpmms.cam.ac.uk/~dc340/RE.pdf) article by D.Conlon.


