Very nice problem! I will use fair instead of honest. Consider a functional graph with $2n$ vertices, where $v_i\rightarrow v_j$ if the proposer of problem $i$ receives problem $j$. The given conditions simply imply there exist a set of $n$ students who didn't receive any problems from the set, so the functional graph is "bipartite", or all cycles are even. Let $(2n-1)!!=\prod\limits_{i=1}^n (2i-1)$ I claim the number of fair tournaments is exactly $(2n-1)!!^2$. We will first prove a lemma. Lemma. $$\sum\limits_{i=0}^{n-1} \frac{\binom{2i}{i}}{4^i} = \frac{(2n-1)!!^2}{(2n-1)!}$$ Proof of Lemma. We will proceed by induction. The base case, $n=1$ is clear. We will assume result for $n$ and prove this for $n+1$. Note $$\frac{(2n+1)!!^2}{(2n+1)!}-\frac{(2n-1)!!^2}{(2n-1)!}=\frac{(2n-1)!!^2}{(2n+1)!} ((2n+1)^2-(2n+1)(2n)) = \frac{(2n-1)!!^2 (2n+1)}{(2n+1)!} $$ $$=\frac{(2n-1)!!^2}{(2n)!}$$ Notice $(2n-1)!!=\frac{(2n)!}{2^nn!}$, so this is actually $\frac{(2n)!^2}{4^n (n!)^2 (2n)!} = \frac{(2n)!}{4^n (n!)^2}=\binom{2n}{n} 4^{-n}$, completing the induction. Now, let $a_n$ be the number of fair selections for $2n$ contestants. If $a_1$ is in a cycle of $2k$ people, then there are $\prod_{i=1}^{2k-1} (2n-i)$ ways to choose the cycle because there are $2n-i$ to choose the $i$th vertex of the cycle for $i<2k$ if the $0$th vertex is $1$. Then we just need any fair selection for $2(n-k)$ people, which has $a_{n-k}$ selections. We proceed by strong induction to show that $a_i=(2i-1)!!^2$. The base case, $i=1$, is clear. Let $a_0=1$ be the number of ways to choose 0 contestants such that the competition is fair. Therefore, $a_n=\sum\limits_{i=0}^{n-1} \frac{(2n-1)!}{(2i)!} a_i = (2n-1)! \sum\limits_{i=0}^{n-1} \frac{(2i-1)!!^2}{(2i)!} = (2n-1)! \sum\limits_{i=0}^{n-1} \binom{2i}{i} 4^{-i}$, so we are done by our lemma.