We shall show in fact that it is impossible for 102 squares. The configuration is a $2\times 2$ square at the top left hand corner, and the remaining squares on the top left-bottom right main diagonal. We show this by induction on a $n\times n$ table. The base case $n=2$ is obvious (all the squares are coloured). For the inductive step, suppose it is possible for some $n>2$. Choose a row containing one marked square. Then the column containing the square does not contain any other marked squares. If this square is not on the main diagonal (i.e. the square is to the right of the diagonal), then we can move the column left so that the square is on the main diagonal. All the other columns are moved right and the validity of this configuration is not affected. Then we can remove this row and column from the board, and apply the inductive hypothesis to obtain a contradiction.