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I tried to prove this by induction.

Base case $n=1$, $5$ vertices. I just drew a pentagon, which has $5$ vertices of degree $2$

Then I assume for $n=k$,$4k+1$ vertices, there is at least one vertex with degree $2n$. The number of edges for this graph is $\dfrac{(4k+1)(4k)}{4}=(4k+1)(k)$

Then for $n=k+1$,$4k+5$ vertices. The number of edges is $\dfrac{(4k+5)(4k+4)}{4}=(4k+5)(k+1)$

The graph with $4k+1$ vertices have $8k+5$ more edges than the graph with $4k$ vertices.

This is where I get stuck.Am I on the right track? How should I proceed from here?

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    In your proof it will be perhaps a bit difficult for you to incorporate self-complimentary property when you go from the $n$ to $n+1$ case. – 2017-01-18
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    Of possible interest, the number of self-complementary graphs for various numbers of nodes: [OEIS A000171](http://oeis.org/A000171) - spectacular numbers, for example there are apparently $293293716992$ self-complementary graphs with $21$ nodes. – 2017-01-19

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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, $\text{deg}(v) = 2n$.

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):

enter image description here

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

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    How did you get $4n$? Are you assuming that each vertex is connected to all the other $4n$ vertices, hence the degree of $4n$? – 2017-01-19
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    The complementary graph 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 the vertex degree from $x$ to $4n-x$ – 2017-01-19
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    I understood the degree sequence part, but why at position $2n+1$? – 2017-01-19
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    That's the middle point of the degree sequence, there are $4n+1$ entries. – 2017-01-19
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Let $G$ be a self-complementary graph of order $4n+1$ and let $\sigma:G\to\overline G$ be an isomorphism. Since $G$ and $\overline G$ have the same vertex set $V,$ $\sigma$ is a permutation of $V,$ an anti-automorphism of $G,$ mapping edges of $G$ to edges of $\overline G$ and vice versa.

Consider the cycle decomposition of the permutation $\sigma.$ (Note that we are using the word "cycle" in the group theory sense, not the graph theory sense!) Each nontrivial cycle of $\sigma$ must be of even length, since edges of $G$ alternate with edges of $\overline G$ as we go around the cycle. Since $|V|$ is even, $\sigma$ must have a $1$-cycle, i.e., a fixed point. (It's clear that the fixed point is unique, but we don't need that.)

Let $v$ be a fixed point of $\sigma.$ Since $\sigma$ is an isomorphism between $G$ and $\overline G,$ we must have $\deg_G(v)=deg_{\overline G}(v).$ Since $\deg_G(v)+\deg_{\overline G}(v)=4n,$ it follows that $\deg_G(v)=\deg_{\overline G}(v)=2n.$

P.S. It's also easy to see that the length of a nontrivial cycle of $\sigma$ is not only even, it's a multiple of $4;$ and if one vertex in a cycle has degree $2n,$ then all vertices in that cycle have degree $2n.$ Since $\sigma$ has a unique fixed point, it follows that the number of vertices of degree $2n$ is congruent to $1$ modulo $4.$

Examples. There are two self-complementary graphs of order $5.$ One of them is $C_5,$ which has five vertices of degree $2;$ the other is homeomorphic to the letter A and has one vertex of degree $2.$