For every boards are two chessboard patterns we can solve for. For example, for a 3x3 board we can have:

  101        010
  010   or   101
  101        010

where 1 is a white square and 0 is a black square.

We can represent any N×N board as a N² vector. So, we can represent one of the chessboard pattern as:

  010
  101 = 010101010 = y
  010

At the beginning of the game, the board is at some random initial pattern.

  110
  111 = 110111010 = y0
  010

The goal of the game is to find the necessary moves to form the chessboard pattern. Hence, we want to solve the equation

  (y - y0) = Mx

where M is a N²×N² matrix representing the effect of every possible moves. Pressing on a tile is equation to addition by 1 modulo 2.

  1 + 1 = 0 (mod 2)
  0 + 1 = 1 (mod 2)

For example, pressing the top left tile is the same as adding this effect vector to the board vector:

  110110000

If we do this for all 9 tiles of the 3x3 board, we find the matrix M to be:

  110110000
  111111000
  011011000
  110110110
  111111111 = M
  011011011
  000110110
  000111111
  000011011

where the i-th row is the effect of pressing the i-th tile. Now, we can just solve for x (assuming M is not singular). Re-using our above example, we find

  (y - y0) = Mx
  (010101010 - 110111010) = Mx
  100010000 = Mx
  111100100 = x   (by Gaussian elimination)

This means we can press the following tiles (in any order) on the board to solve it:

  111
  100
  100

If you are interested, I wrote some quick demo code that shows how this could be implemented in Python: https://gist.github.com/avassalotti/7452318c9e9b8dec637bc7c0...

This is a very nice exposition of the strategy, but I still don't see how it answers my more specific question, about whether you can solve the problem row by row (is the grid).

Sorry, it turns that was misunderstanding from my part. It turns out we cannot solve a board row by row without revisiting the previous rows. The board below is a counterexample:

  101
  010
  100

The first and second rows are solved. However we cannot solve the last row without re-doing the first and second rows. The two solutions of this board shows this:

  solution #1:
  110
  100
  000

  solution #2:
  110
  110
  000

Actually, seeing my mistake made me challenge my assumption that a singular matrix over the reals might not be over integers modulo 2. This is likely wrong too. I don't know much about abstract algebra (and I am not a mathematician). Wikipedia (https://en.wikipedia.org/wiki/Determinant#Square_matrices_ov...) states "the reduction modulo m of the determinant of such a matrix is equal to the determinant of the matrix reduced modulo m."

The move matrix M for all boards of size 3n + 2 appears to be singular. This means these boards may have no solution or a large number of solutions.

Gotcha, thanks.

FYI, a singular matrix taking values in {0,1} that is singular over R if and only if it is also singular over Z/2, for exactly the reason you provide, that the determinant commutes with modding by 2.