Let $P_k$ be the person who shook $k$ hands. $P_{2n-2}$ shook hands with everyone but his or her partner, so $P_0$ must have been the partner. Set them aside, leaving $P_1,\dots,P_{2n-3}$ and the mathematician. Each of these remaining people shook hands with $P_{2n-2}$, so within the group that remains each shook one hand fewer: $P_k$ shook $k-1$ hands for $k=1,\dots,2n-3$. By the same reasoning (or by induction) $P_1$ and $P_{2n-3}$ must be a couple. In general we must have $P_k$ and $P_{2n-2-k}$ forming a couple for $k=0,\dots,n-2$. In particular, $P_{n-2}$ and $P_n$ are a couple. This leaves $P_{n-1}$ to be the mathematician’s husband: he shook $n-1$ hands.
In graph-theoretic terms we have a graph $G_n$ with vertices $v_k$ for $k=0,\dots,2n-2$ such that $\deg v_k=k$, and we have an additional vertex $v$ corresponding to the mathematician. The graph is simple (no loops or multiple edges), and it is known that the vertices can be partitioned into $n$ pairs whose members are not adjacent. We wish to show that $v$ is paired with $v_{n-1}$. This is clearly the case when $n=1$, so assume that $n>1$.
Vertex $c_{2n-2}$ is adjacent to every vertex but itself and the one paired with it, so every vertex but the one paired with it has positive degree; thus, $\{v_0,v_{2n-2}\}$ must form a pair. Remove $v_0$, $v_{2n-2}$, and all adjacent edges, and you have a graph $G_{n-1}$ on $2(n-1)$ vertices with mutatis mutandis the same properties. By the obvious induction hypothesis vertex $v$ is paired with the vertex of degree $n-2$ in $G_{n-1}$, which is $v_{n-1}$ in $G_n$, as desired.