[Proof] Consider the sum of the numbers filled in each row and each column, and let the smallest of these $2m$ numbers be denoted as $p$. Without loss of generality, assume $p$ is the sum of the numbers in the first row. If $p \geqslant \frac{m}{2}$, then the conclusion to be proven is obviously true. Suppose $p < \frac{m}{2}$. By estimating the sum of the numbers filled in each column, we can see that the total sum $S$ of all numbers filled in the table satisfies
$$\begin{aligned} S & \geqslant (m-p) k + (m-k) p = m(k + p) - 2 p k \\ & = m(k + p) - \frac{1}{2}(p + k)^{2} + \frac{1}{2}(p - k)^{2} \\ = & \frac{m^{2}}{2} + \frac{1}{2}(p - k)^{2} - \frac{1}{2}(p + k - m)^{2} = \frac{m^{2}}{2} + \frac{1}{2}(2 p - m)(m \\ & - 2 k) \\ & > \frac{m^{2}}{2} \end{aligned}$$