Annual Harvard-MIT Mathematics Tournament

Solution:

Answer: $372$

Let $s_{n}$ be the minimum possible sum for an $n \times n$ matrix. Then, we note that increasing it by adding row $n+1$ and column $n+1$ gives $2n+1$ additional entries, each of which has minimal size at least $n+1$. Consequently, we obtain
$$s_{n+1} = s_{n} + (2n+1)(n+1) = s_{n} + 2n^{2} + 3n + 1.$$
Since $s_{0} = 0$, we get that
$$s_{8} = 2\left(7^{2} + \ldots + 0^{2}\right) + 3(7 + \ldots + 0) + 8 = 372.$$