Suppose two players play the following game on a $m$ by $n$ rectangle. Alternatingly they have to make a cross in some empty $1\times 1$ square. They are not allowed to make a cross next to another cross (Diagonally is OK, not just right next to each other). The player who places the last cross wins.
Now the question is for which $m,n$ does the starting player have a winning strategy?
At first I thought this might be just a nice exercise. So I had a look at 1 by $n$ rectangles first. A computer programme computed that the first player does not have a winning strategy for $n=4,8,14,20,24,28,34,38,42,54,58,62,72,76,88,92,96,106,110$ (for $n\le 110$).
I do not see any pattern in these numbers. So the answer might not be so easy.
The starting player can always win for odd $n$. He just places a cross in the middle and mirrors all the moves of the second player.