Make all cells of the square be the same color $200 \times 200$ square is colored in chess order. 
In each move we can take any $2 \times 3$  rectangle and change the color of all its cells. 
Can we make all cells of the square be the same color ?

I still do not have an idea. 
Is this construction useful, square with every $2 \times 3$  rectangle contains 2 black cells ?

 A: Suppose you managed to make all cells black. Rotate  board so that left-upper corner is black. Now place in each cell with coordinates $(i,j)$ remainder of $(i-j)$ modulo $3$:
$$
\begin{array}{|c|c|c|c|c}
\hline
\bf0 & 1 & \bf2 & 0 &\ldots\\ 
\hline
2 & \bf0 & 1 & \bf2 &\ldots\\
\hline
\bf1 & 2 & \bf0 & 1 &\ldots \\
\hline
0 & \bf1 & 2 &\bf 0 &\ldots\\ 
\hline
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
$$
Denote number of black $1$'s as $B$. In beginning $B$ is even, and when all cells are black, $B$ is odd. However, if you repaint all cells in $2\times 3$ rectangle, parity of $B$ remains the same, since there is always two $1$'s in rectangle. That's a contradiction. 
A: I almost found a solution, if You mean 2x3 and 3x2 blocks (in another words if we can rotate our block by 90 degrees). I show how to make it for 5x5, then 5x5 blocks generate 200x200 square.  
10101
01010
10101
01010
10101  
11001
11010
00100
01011
10011  
11111
11011
10101
11011
11111  
11111
10001
11111
10001
11111  
11111
11111
11111
11111
11111  
Try to resolve my turns in each phase. Hint: I only used double turns of 2x3/3x2 inverses. One of the main facts is that when we apply two inverses, the intersection of these inverses will not be inversed. I used double inverses with keeping 2x2 squares and changing other four points which are not in the intersection.
Edit:
As it was told in comments, there is a trouble that color in adjacent blocks eill be inversed. May be we can bypass this..
