The answer is $n-m+1$. We will denote black squares by $1$ written in them and white with $0$ and thus the board as a matrix $X$. Coloring a column red will correspond to vanishing it. To show that anything greater than $n-m+1$ doesn't work. Consider the matrix with $x_{ij} = 1 \iff i=j$. It is easy to see that vanishing the last $n-m+1$ columns works but vanishing anything more than that won't work (at least $2$ rows are left with only zeros). To show that $n-m+1$ indeed works, we need to show that the difference in the rows can be captured in $m$ columns which I think is induction