Consider the square board to be of dimensions $N\times N$, with $N=n^2+1$, filled with the numbers $\{1,2,\ldots,N\}$, each used $N$ times. Denote by $r_i$ the number of rows containing the number $i$, and by $c_i$ the number of columns containing the number $j$, for all $1\leq i\leq N$. Clearly the total number of apparitions of the number $i$ is at most $r_ic_i$, hence $N\leq r_ic_i \leq \left (\dfrac {r_i+c_i} {2} \right )^2$. Denote by $k_i = r_i + c_i$ the number of rows or columns containing the number $i$. Thus $k_i^2 \geq 4N = 4(n^2+1) = (2n)^2 + 4$, so $k_i \geq 2n+1$. Summing over $i$ yields $\sum_{i=1}^N k_i \geq N(2n+1)$. On the other hand, if each row and each column would only contain at most $n$ different numbers, then each of the $2N$ rows and columns is counted at most $n$ times within the above sum, therefore $\sum_{i=1}^N k_i \leq 2Nn$. Since these two inequalities cannot occur simultaneously, it means at least one row or column contains at least $n+1$ different numbers.