Let $(i,j)$ denote the square in both the $i$-th row and $j$th column. Then the idea is to colour $(1,1)$ and $(1,2)$ black,$(1,2) $and $(2,2)$ white, $(3,1)$ and $(3,2)$ black ....... $(13,1)$ and $(13,2)$ black, $(1,3)$ and $(1,4)$ white, $(2,3)$ and $(2,4)$ black etc for the rest of that double column and there after $(i,j)$ has the same colour as $(i,j-4)$. I'll edit my post later to add a picture. From this colouring clearly any 4x1 block will cover two white squares and two black squares. There are 85 black squares and 83 white squares (recalling that the central white square is removed). Contradiction - $2\nmid 83,85$.

As promised, a diagram here or:

Image

This is an nice problem and could be dn easily by introducing complex number here