Possibilities in board houses Let $n> 1$ be an integer and consider a board $n$ × $n$, where some of $n ^ 2$ houses were painted black, and the rest were painted white. Prove that we can choose one of $n ^ 2$ board houses, from so that when we completely remove the row and the column that contains it, there will be a different number of squares black and white squares, among the $(n - 1) ^ 2$ remaining houses
What i tried : Let n be an even number, $(n-1) ^ 2 $ is odd, absurd (why?)
Let n be odd. Let's try to make the above statement always fail, that is, they are always equal to the remaining $ (n-1) ^ 2 $ boxes. Choose a random house and one next to it.
 A: As you noted, the case that $n$ is even is trivial (not absurd, IMO), because the remaining $(n-1)^2$ houses are an odd number, so you can choose any house, remove the row/column it is in and get a different number of black/white houses in the remaing houses. This is a much stronger result that what you need to prove. For odd $n$ this doesn't work, though.
So let's assume $n$ is odd from now on. One possible idea in this kind of problems is to consider all $n^2$ different scenarios (where you select a house and then determine the remaining houses as described in the problem) 'at the same time'.
If we have one such scenario where the number of black/white houses are different, we are done. If not, that means it is the same in each scenario. So when we add the number of black houses in each scenario, we get the same number $S$ as when we add the number of white houses in each scenario. 
OTOH, during this adding up, we count each house exactly $(n-1)^2$ times in those $n^2$ scenarios (we don't count it if in the given scenario we selected a house in the same row or column, which accounts for $2n-1$ cases).
But this means that $S$ is both $(n-1)^2$ times the number of white houses on the whole board and $(n-1)^2$ times the number of black houses on the board, which means the number of black houses on the board is equal to the number of white houses on the board.
Now that is impossible, as $n$ was assumed odd, so the black/white houses can't have the same number.
