1) Let $B_n$ be the board after $n$ moves. Consider numbers $\mod 2$. If there are $m$ red squares there are no more than $2^m$ - a finite number - of states the board can be in. 2) The transformation, $T(B_n)=B_{n+1}$ is invertible. Call $(i,j)$ of order $1$ if $(i,j)$ is the only red square in $S_{i,j}$. Inductively, call $(i,j)$ order $n+1$ if there exists $(k,\ell)\in S_{i,j}$ where $(k,\ell)$ is order $n$ but there is no square of order $\ge n+1$ in $S_{i,j}$. The order of a square remains the same each move. Now to invert $T$ start at a square of order $1$. We know these squares are always $1$, so we know their previous state. Next move to squares of order $2$, these squares only contain squares of order $1$, and we know those squares previous state hence we can work out the previous state of order $2$ squares (in $mod 2$). Continuing in this manner we can invert $T$. From (1) and (2) it follows that the sequence $B_0,B_1,B_2,..$ is purely periodic. Now consider the board in which every red square had the number $1$. In one move if $S_{i,j}$ has an even number of red squares then $(i,j)$ becomes $0$ and if $S_{i,j}$ has an odd number of squares then $(i,j)$ becomes $1$. But this is the initial state. Since $B_n$ is purely periodic we must reach the state where every number is $1$ (i.e. odd) in a finite number of moves (in fact, it must be less than $2^{m-1}-1$ moves when there are $m$ red squares)

Is this true that the steps required to make all squire odd is a power of 2 ?

Roozbeh wrote: Is this true that the steps required to make all squire odd is a power of 2 ? Yes it is. It can be proven by induction.