In an $n\times n$ chessboard, $n>2$, $M$ squares have an ant in their center. Each ant starts moving with the same speed parallel to some edge of the chessboard. When they reach the center of the next square, they turn $90^{\circ}$ to the left or right and continue moving in that direction. The ants always choose their direction of turning so that they won’t fall off the edge of the board.
What is the minimum value of $M$ so that we can guarantee that eventually there will be two ants on the same square?
This is a variant of IMO Shortlist 2011 C5. I feel that the pigeonhole principle will be useful here, but I'm not sure how to apply it.
For $n$ even, it is possible to have $M=n^2$ ants without them ever occupying the same square, as per Max Freiburghaus' example in the comments below. For odd $n$, is $M=(n-1)^2+1$ the answer?
Note. After two moves each ant appears in a cell that has exactly one common corner with his current cell.
Suppose that ant can move infinitely long with no collision that is two ants in the same cell.
Let us enumerate both rows and columns one after another with numbers $1, \ldots, n$. If $n = 2k + 1$ then there are $k^2$ cells that have both coordinates even and these cells can contain up to $k^2$ ants. By note above we can deduce that there are also at most $k^2$ ants on cells with both coordinates odd. After one move each other ant (on a cell with an odd sum of coordinates) goes to a cell with even sum of coordinates that is one of considered above. This means that number of such ants is at most $2k^2$, too, that gives at most $4k^2 = (n - 1)^2$ ants in total.
Thus for $M = (n - 1)^2 + 1$ there definitely will be a collision.