We color the cells a $4n\times 4n$ board ($n\geq 1$) in black and white. What is the maximum number of "rectangles", i.e. four cells that together form a rectangle with sides parallel to the sides of the board, such that the two cells in the opposite corner have one color and the other two cells have the other color?
For a chessboard coloring, the number is $(2n)^4$. For a coloring that puts black on the top-left $2n\times 2n$ and the bottom-right $2n\times 2n$, the number is also $(2n)^4$.
For $n=1$, $(2n)^4$ is the answer. But the argument uses case division. For arbitrary $n$ we can say that in the two examples above where the bound is achieved, each square is a corner of $(2n)^2$ different good rectangles, and if there are more than $(2n)^4$ good rectangles, some square must be in at least $(2n)^2+1$ good rectangles.