Let $b_i$ be the number of blue cells in the $i$-th row. Then $b_1+b_2+\dots+b_n=k$ and the number of red cells in the $i$-th row is $n-b_i$. So the sum of squares of the number of blue cells in each row minus the sum of squares of the number of red cells in each row is $$\sum_{i=1}^n (b_i^2 - (n-b_i)^2) = \sum_{i=1}^n (2nb_i - n^2) = 2nk-n^3 = n(2k-n^2).$$Similarly, the same can be said for columns. So $S_B-S_R=2n(2k-n^2)$ and $n(2k-n^2)=25$. So $(n, k) = (1, 13), (5, 15), (25, 313)$. But the number of blue cells must be at most the number of total cells, so $k < n^2$. Thus, $(n, k)$ cannot be $(1, 13)$. Note that $S_B - S_R = 50$ can be attained by coloring $15$ cells of a $5 \times 5$ grid blue or by coloring $313$ cells of a $25 \times 25$ grid blue. Therefore, $\boxed{15}$ and $\boxed{313}$ are all the possible values of $k$.