The idea is to double count the triplets (column, column, row), where the two columns coincide in the position which represents the row. If the number of triplets is $T$, then the idea is to prove $\frac{m(m-1)}{2}k \geq T \geq \frac {\frac{m^2}{3}-m}{2}n=\frac{(m^2-3m)n}{6}$ where the first one is due to the problem statement, the second one due to the following: if there are $n_i$ cells of each color in a fixed row, then the number of pairs of columns which coincide in position of this row is $\sum \frac{n_i(n_i-1)}{2}$ and you can find lower bound of this by CS. The rest of the problem is to see equality holds.