We will for simplicity and clarity of presentation use some basic graph theoretic terms, this is in no way essential. We represent the situation by a directed graph (abbr. digraph) $G(V, E)$ where each vertex $v \in V (G)$ represents a mathematician and each edge $e \in E(G)$ represents an admiration relation. Given $v \in V (G)$ we define out-degree of $v$ denoted $o(v)$ as the number of edges starting in $v$ (so the number of mathematicians $v$ admires) and in-degree $i(v)$ as the number of edges ending in $v$ (so the number of mathematicians who admire $v$). Given $X \subseteq V$ by $G(X)$ we denote the induced subgraph (a graph with vertex set $X$ and edges inherited from $G$). We say that a digraph is a $k$-digraph if for every $v\in V (G)$ we have $i(v) \ge k$ or $o(v) \ge k$. So the question can be reformulated as: Given $G$ is a $3k + 1$-digraph we can split its vertices into $2$ vertex disjoint classes such that each induced subgraph on class is a $k$-digraph. We call a subset $X$ of vertices of $G$, $k$-tight if for any $Y \subseteq X$ we have a vertex $v \in Y$ such that $i_{G(Y )}(v) \le k$ and $o_{G(Y )}(v) \le k$. A partition of $V$ , $(A_1, A_2)$ is feasible if $A_1$ is $k$-tight and $A_2$ is $k$-tight. We first assume there are no feasible partitions. In this case consider a minimal size subset $A_1 \subseteq V (G)$ subject to $G(A_1)$ being a $k$-digraph, we define $A_2 \equiv V (G) - A_1$. Given a subset $X \subset A_1$, $G(X)$ is not a $k$-digraph so there is a vertex $v \in X$ such that $o_{G(X)}(v) < k$ and $i_{G(X)}(v) < k$ which shows that any proper subset of $A_1$ satisfies the condition of $k$-tightness. For the case of $X \equiv A_1$, by removing any vertex $v \in A_1$ the graph $G' \equiv G(A_1 - \{v\})$, by minimality assumption on $A_1$, must contain a vertex $w$ such that $o_{G'} (w) < k$ and $i_{G'} (w) < k$ so as there is only one extra vertex in $G(A_1)$, namely $v$, so $o_{G(A_1)}(w) \le k, i_{G(A_1)}(w) \le k$. In particular this shows $A_1$ is $k$-tight. This implies $A_2$ is not $k$-tight by our assumption so there exists an $A'_2 \subseteq A_2$ such that $A'_2$ is a $k + 1$-digraph. Now applying the following proposition to extend the pair $(A_1, A'_2)$ to a full partition which satisfies the conditions of the problem. Given disjoint subsets $A, B \subseteq V (G)$ we say $(A, B)$ is a solution pair if both $G(A)$ and $G(B)$ are $k$-digraphs. Proposition: If a $2k+ 1$-digraph $G$ admits a solution pair it admits a partition with both induced graphs of both classes being $k$-digraphs. Proof. Take a maximal solution pair $(A, B)$, the condition in the lemma guaranteeing it exists. Let $C = V (G)-(A\cup B)$, if C is empty we are done so assume $|C| > 0$. By our assumption $(A, B \cup C)$ is not a solution pair so there is some $x \in C$ such that $o_{G(B\cup C)}(x), i_{G(B\cup C)}(x) < k$ so as $G$ is $2k + 1$ digraph $i_G(x) \ge 2k + 1$ or $o_G(x) \ge 2k + 1$ so either $o_{G(A\cup \{x\})}(x) > k + 1$ or $i_{G(A\cup \{x\})}(x) > k + 1$ so in particular $(A \cup \{x\}, B)$ is a solution pair contradicting maximality and completing our argument. $\square$ Hence we are left with the case in which we have at least one feasible partition. We pick the feasible partition $(A, B)$ maximizing $w(A, B)=|E(G(A))| + |E(G(B))|$. The fact that $A$ is $k$-tight implies there is an $x$ with $o_{G(A)}(x) \le k$, $i_{G(A)}(x) \le k$ so $x$ needs to have at least $k + 1$ edges in or out of $B$, so $|B| \ge k + 1$ and by symmetry $|A| \ge k + 1$. We now prove that there exist an $X \subseteq A$ such that $G(X)$ is a $k$-digraph, by contradiction. Assuming the opposite we notice that for any $x \in B$, $B - \{x\}$ is still $k$-tight while $B$ being $k$-tight implies there is an $x \in B$ such that $o_{G(B)}(x) \le k,i_{G(B)}(x) \le k$ so for this $x$ we have $A\cup \{x\}$ is also $k$-tight. Hence, for $A' = A\cup \{x\}$ and $B' = B - \{x\}$, $(A', B')$ is a feasible partition. Considering the change in edges which moving $x$ causes, we have $w(A', B')-w(A, B) \ge 3k+1-k-k-k = 1$, as we know $i_G(x) \ge 3k + 1$ or $o_G(x) \ge 3k + 1$, so moving $x$ from $B$ to $A$ increases number of edges in $A$ by at least $3k + 1 - k$, while the choice of $x$ in $B$ means we lose at most $k + k$ edges in B. This is a contradiction to maximality of $(A, B)$. Analogously we can find $Y \subseteq B$ with $G(Y )$ a $k$-digraph. Now applying the above proposition yet again we are done. $\blacksquare$