Prove every self-complementary graph with $4n+1$ vertices has at least one vertex of degree $2n$

The complementary graph to $G$ adds all edges that are missing from the complete graph $K_{4n+1}$, and removes all existing edges. In $K_{4n+1}$ every vertex has $4n$ edges, so the complementing process changes each vertex degree from $x$ to $4n-x$.

The self-complementary graph must have the same degree sequence as its complement, and each vertex has its degree change from $\text{deg}(v)$ to $4n{-}\text{deg}(v)$ in the switch to the complement - so there must also be a vertex of that degree in the original graph to switch back. This corresponds to switching between an early position in the degree sequence and a late one.

So at the middle position $2n{+}1$ in the degree sequence, we must have a vertex that switches value with itself; that is, $\text{deg}(v) = 4n{-}\text{deg}(v)$. Which means for at least that vertex, $\text{deg}(v) = 2n$.

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As an aside, you might also like to consider the other $5$-vertex self-complementary graph (apart from the $5$-cycle) in any proof you attempt: (from http://mathworld.wolfram.com/Self-ComplementaryGraph.html):

Here $n=1$ of course and the middle position of the degree sequence is $3$ - and we see the sequence of $(3,3,\color{red}{2},1,1)$.