I need help with this graph proof please Let $G$ be a self-complementary graph in where $n = 4k+1$ . Prove that there is an odd number of vertices of degree $(n-1)/2$.

I dont even know where to start with this. I really need help.

Thanks!
As it is isomorphic to its complement we can say that 

$$\sum_{i=1}^n deg(v_i)=\sum_{i=1}^n 4k-deg(v_i) $$ 

Then we have $$2\sum_{i=1}^n deg(v_i)=\sum_{i=1}^n 4k=4k(4k+1) $$

$$\sum_{i=1}^n deg(v_i)=2k(4k+1)$$

Notice that if $r=deg(v_i)$ then $4k-r$ is also a degree of another vertex. And if $r\neq 2k$ ,it is clear that it is another vertex. Set $s$ be number of vertex with degree $2k$ and $l$ be the number of vertex with degree less than $2k$. 

$$s2k+l(4k-r+r)=2k(4k+1)$$
$$s+2l=4k+1$$
$$s\equiv1 \text{ } mod(2)$$
Hint: $v$ has degree $\frac{n-1}{2}$ in $G$ if and only if $v$ has degree $\frac{n-1}{2}$ in its complement...

Can you  pair the vertices which don't have degree $\frac{n-1}{2}$?