I calculated that Anne can choose 511 ways (she either chooses a square or not, and there are nine of them, so 2^9 possibilities, but she can't choose none of them). I also noticed if the diagonally opposing squares of a 2x2 square are chosen, and the other two are not, then it is a solution for Anne, because she can choose the other two squares of the 2x2 square and still tell the same number. This also holds true for 2x3 rectangles and the corners of the 3x3 square.

I tried to figure it out combinatoricly using this fact, but I just couldn't find a way to do so. Any help or idea is appreciated.