Harvard-MIT Mathematics Tournament

Solution:

Let $c_{i, j}$ denote the number of students correctly answering questions $i$ and $j$ ($1 \leq i < j \leq 10$), and let $w_{i, j}$ denote the number of students getting both questions wrong. An individual student answers $k$ questions correctly and $10-k$ questions incorrectly. This student answers $\binom{k}{2}$ pairs of questions correctly and $\binom{10-k}{2}$ pairs of questions incorrectly. Now observe that
$$\binom{k}{2} + \binom{10-k}{2} = k^2 - 10k + 45 = (k-5)^2 + 20 \geq 20$$
Therefore,
$$\sum_{1 \leq i < j \leq 10} c_{i, j} + w_{i, j} \geq 20n$$
Now if the performance is not laughable, then $c_{i, j} \leq 56$ and $w_{i, j} \leq 56$ for all $1 \leq i < j \leq 10$. Observe that there are $2 \binom{10}{2} = 90$ of these variables. Hence, in a boring performance,
$$20n \leq \sum_{1 \leq i < j \leq 10} c_{i, j} + w_{i, j} \leq 90 \cdot 56 = 5040$$
or $n \leq 252$. In particular this implies that if $n \geq 253$, the performance is laughable. This is the best bound because $\binom{10}{5} = 252$, and if each of 252 students correctly answers a different 5-element subset of the 10 questions, then $c_{i, j} = w_{i, j} = 56$ for all $1 \leq i < j \leq 10$.