If the $i$-th row of the grid has $b_i$ blue cells, then the contribution to $S_B - S_R$
from this row is $b_i^2 - (n - b_i)^2 = 2nb_i - n^2$. Adding these contributions over
all rows yields $2nk - n^3$ and similarly the columns also contribute $2nk - n^3$;
thus we have
$$S_B - S_R = 2(2nk - n^3) = 2n(2k - n^2).$$
Next, denote the number of red cells by $\ell$, and suppose that there are $t$ more
blue cells than red cells. Then $k+\ell = n^2$ and $k-\ell = t$, which implies $n^2 \ge t$.
When we add these two equations we obtain $2k = t+n^2$, hence $t = 2k-n^2$
and
$$S_B - S_R = 2n(2k - n^2) = 2nt.$$
Since we are given that $S_B - S_R = 50$, this implies that $nt = 25$ and so
$(n,t) = (5,5)$ or $(n,t) = (25,1)$. Note that $(n,t) = (1,25)$ is not possible as
$n^2 \ge t$. Using $2k = t+n^2$, in the first case, we get $k = 15$ and in the second
case, $k = 313$.
The equation $S_B - S_R = 2n(2k - n^2)$ shows that $S_B - S_R$ only depends on
the size of the grid and the number of blue cells, but not on the position
of the blue cells. Therefore, making any 15 of the 25 cells of a 5 × 5 grid
blue gives a colouring which satisfy the conditions of the problem for $k = 15$.
Also, making any 313 of the 625 cells of a 25 × 25 grid blue gives a colouring
which satisfy the conditions of the problem for $k = 313$.