Let $n \ge 2$ be an integer. In each cell of a $4n \times 4n$ table we write the sum of the cell row index and the cell column index. Initially, no cell is colored. A move consists of choosing two cells which are not colored and coloring one of them in red and one of them in blue.
Show that, however Alex perfors $n^2$ moves, Jane can afterwards perform a number of moves (eventually none) after which the sum of the numbers written in the red cells is the same as the sum of the numbers written in the blue ones.
This doesn't seem to bad, yet I feel like finding a rigourous construction that always works is not easy.
Here is a sketch:
Anyway, if you let $A$ be the sum of red cells Alex made and $B$ the blue ones, then $|A-B| < (8n-2)n$ obviously, as $8n$ is the highest value in a cell and $2$ the lowest. Now you want to prove that in the other $14n^2$ cells, there exists a partition of sets $U$ and $V$ over the other $14n^2$ cells (colours red and blue) such that $|U-V| = |A-B|$ with cardinality $U$ and $V$ equal. Notice that Jane can make at most $7n^2$ moves and thus it seems obvious that she can easily get a sum equal to $|A-B|$, but finding a rigourous construction without thinking about size argument isn't that obvious I'd say. I think the main idea is to show that you can make take plenty of cells with difference $1$ and put them in $U$ and $V$ respectively to get to the actual value of $|A-B|$ precisely, and then the rest of the cells you want big differences to get in the area of $|A-B|$, but just under it. This should definitely be doable and will lead to a proof, but I just don't have the motivation to carry this out xd]