without loss of generality we can bring the top-left vertice to 1x1, so that the vertices of the frame are (going counter-clockwise): $(1, 1); (1, m); (n, m); (n, 1)$ so, supose the square (1, 1) is painted black the sum of the differences between white and black squares will be: colunm (1, 1): $(2-1)+(4-3)+(6-5)+...+(m-1-[m-2])=\frac{m-1}{2}$ row (1, m): $(2m-m)+(4m-3m)+(6m-5m)+...+[(n-1)m-(n-2)m]=\frac{m(n-1)}{2}$ colunm (m, n): $n(m-1-m)+n(m-3-[m-1])+n(m-5-[m-4]+...+2n-3n=-\frac{n(m-1)}{2}$ row (n, 1): $[n-1-n]+[(n-3)-(n-2)]+[(n-5)-(n-4)]+...+3-4+1=-\frac{n-1}{2}$ the sum of the sum of the differences is: $\frac{m-1}{2}-\frac{n(m-1)}{2}+\frac{m(n-1)}{2}-\frac{n-1}{2}=\frac{(1-n)(m-1)}{2}+\frac{(n-1)(m-1)}{2}=0$