Since nobody wants to give a solution, Notice that the number of white squares and black squares does not change in every movement (mod 4), since this 4 must divide 2n^2 since there are half white and half black, this only occurs when "n" is even. A nice example of this is when "n" equals 4.

Your solution is wrong in all ways, as the claimed invariant doesn't work (the numbers of white squares and black squares mod 4 change when a 2x2 square applied has 3 whites and 1 black or vice versa; they don't change mod 2 instead), and you didn't show that all even $n$ works. Proof that all odd $n$ doesn't work: Take any row. Initially there are $n$ white squares in this row, which is odd. Every move either doesn't touch this row, which means the number of white squares obviously remains the same, or otherwise touches exactly two squares in this row, which means the number of white squares remains the same mod 2. Initially there are an odd number of white squares but at the end there are an even number of white squares in this row, impossible. Algorithm to solve all even $n$: I will give an algorithm that solves a $4 \times 4$ board without changing the states of the others. This way, it's trivial to divide a $2n \times 2n$ board to several $4 \times 4$ boards, and apply this algorithm to each board. We notate each move according to the center of the $2 \times 2$ square that the move is applied to. Suppose our board is as follows: 0|1|0|1 -+-+-+- 1|0|1|0 -+-+-+- 0|1|0|1 -+-+-+- 1|0|1|0 We will only apply moves on those crosses. This way, it's obvious that no square outside this board is changed. The required move is as follows: .|.|.|. -+-O-O- .|.|.|. -O-+-O- .|.|.|. -O-O-+- .|.|.|. It's easy to verify that when the moves are done at the O's, precisely the 1's in the board becomes 0's and so are of the same color. So we're done; all even $n$ (and only those) work.

Oh sorry anyways here is my CORRECT SOLUTION, or I hope so. Actually I will only present the $n$ odd part because the even part is trivial. Color the $2n \times 2n$ grid as the following, (a cell that is colored with $i$ and with black, is said to be $i$-black). 12341234...123412 34123412...341234 ......... 34123412...341234 12341234...123412 There are $n^2-n$ black cells of the color 1, $n^2-n$ black cells of the color 4, and $n$ black cells of the color 2, $n$ black cells of the color 3. In each movement we change the parity of every color of the 4, (1-black, 2-black, 3-black, and 4-black), since a movement changes exactly one of each, (1, 2, 3, 4), we know that $n^2-n$ is even and $n$ is odd,so there always will be 2 odds and 2 evens; because the colors 1-black, 4-black are changed in 1 mod 2, and the colors 2-black, 3-black also are changed in 1 mod 2. Following this, the 4 colors (1-black, 2-black, 3-black, and 4-black), will never share the same parity so there will never be 0 black cells or white cells. So $n$ odd this is impossible. (Please check if I'm wrong).

n must be an even number. It is easy to find the solution for n=2. So for n=2k, we can find the solution. So we only need to prove that n=odd can not satisfy. Mark every black cell by 1 and each white cell by 0. Consider the sum of each line. Because n is an odd number, the sum of each line is odd. The final goal is to make all the cells be the same color, so finally, the sum of each line is even. Consider each movement, it changes the color of two cells in one line. 2 black->2 white; sum-2 2 white->2 black; sum+2 1 black 1 white->1 white 1 black; no change in sum of this line. So the movement does not change parity of the sum of each line, so it can not achieve the goal that each cell in one line is the same color.