Counting edges easily shows that if $n$ is congruent to 2 or 3 modulo 4, there is no self-complementary graph on $n$ vertices. Is the converse true?
What I know: Paley graphs are self-complementary, so if $n$ is congruent to 1 mod 4 and is a prime power, then there is a self-complementary graph on $n$ vertices. Also, the set of $n$ such that there is a self-complementary graph on $n$ vertices is closed under products. Together (since there is a self-complementary graph on 4 vertices) these imply that if the prime factorization of $n$ has even number of 2s and an even number of every prime congruent to 3 modulo 4 then there is a self-complementary graph on $n$ vertices.
Best Answer
Note that if $G$ be a self-complementary graph with $n$ vertices, then the following gives a self-complementary graph $H$ on $n+4$ vertices: Let $H$ be the graph obtained by adding 4 new vertices $\{a,b,c,d\}$ to $G$, with edges $(a,b),(b,c),(c,d), (a,x), (d, x)$ for any vertex $x$ of $G$.
Why $H$ is self-complementary? As $G$ is self-complementary, there is a permutation $f:V(G)\to V(G)$ such that, for any $u,v$ distinct vertices of $G$, $(u,v)$ is an edge in $G$ if and only if $(f(u),f(v))$ is not an edge in $G$. We can define $g:V(H) \to V(H)$ by $g(a)=b; g(b)=d; g(c)=a; g(d)=c$ and $\left. g\right|_{V(G)} = f$, and it is easy to see that $g$ is an isomorphism between $H$ and its complement.
As there are self-complementary graphs on $1$ vertex and on $4$ vertices, it follows that, for any $n=0,1\quad (mod 4)$ there is a self-complementary graph on $n$ vertices.