Suppose that $21$ girls and $21$ boys enter a mathematics competition. Furthermore, suppose that each entrant solves at most $6$ questions, and for every boy-girl pair, there is at least $1$ question that they both solved. Show that there is a question that was solved by at $3$ three girls and at leat $3$ boys.
My question is, why is the number $3$ special here, for example, I tried proving this question in the following way. Fix a girl $g$, this girl could've at most solved $6$ questions $q_1,\dots,q_6$. For the sake of contradiction, assume Each question was solved by at most 2 boys or at most two girls. We first consider if there is at most $2$ boys, then for each of the $6$ questions $g$ answered, there could be at most $2$ pairs involving her and $2$ other boys that solved this question, so for all her $6$ questions, there could be at most $12$ pairs that involve her and other boys. If we sum up this number over all $g$, ($21$ in total), we get the number of pairs is at most $21\times 6 \times 2=252$. This is a contradiction since there are $21^2=441$ pairs of boy-girls, the second case for at most $2$ girls also goes the same by symmetry, so in both cases we have a contradiction.
My question: Why is the number $3$ special here? If it would've been replaced by $4$, then our last equation would be $21 \times 6 \times 3=378$ which is still less than $441$, am I missing something here?