Taiwan IMO Selection Camp, 2021

Let $c_i$ and $r_i$ be respectively the number of columns and rows containing the $i$-th species. Since the intersections of these columns and rows must contain all 110 guinea pigs of species $i$, we have $c_i \times d_i \ge 110$, hence $c_i + d_i \ge 11 + 10 = 21$. This implies $\sum_i (c_i + d_i) \ge 110 \times 21 = 2310$. Also, there are 220 rows and columns in total, so by the pigeonhole principle, at least one of them contains $\left[\frac{2310}{220}\right]+1=11$ different species. Finally, consider the construction where the position $(x, y)$ is a guinea pig of species $(11\left[\frac{x}{11}\right] + \left[\frac{y}{11}\right] + 1)$, and then we know that $n=11$ is indeed the answer sought.