A directed graph whose underlying undirected graph is complete is called a *tournament*. Let us call a (finite) directed graph *balanced* if every vertex has as many incoming as outgoing edges. The question is: Have balanced tournaments been classified? (The weakest form of "classify" might be: given $n$, determine the number of balanced tournaments on $n$ vertices up to isomorphism.)

Here are some elementary observations:

for even $n$ there is no balanced tournament on $n$ vertices.

for odd $n$ there is a standard balanced tournament on $n$ vertices: take as vertex set $\{0,\ldots,n-1\}$ and include an arrow from $i$ to $i+j \mod n$ for $1 \le j \le (n-1)/2$. (The automorphism group is cyclic of order $n$.)

for $n=1$, $n=3$, and $n=5$ the only balanced tournament is the standard one.

for $n=7$ there is a non-standard balanced tournament: to construct it invert an appropriate $3$-cycle in the standard b.t., to see that it is non-standard look at the "out-link" of an appropriate vertex. (The automorphism group is trivial.)

One can take a sort of wreath product to construct examples whose automorphism groups are abelian, non-cyclic. There are other constructions to produce examples.

The motivation for this question is simply the following: the b.t. on $3$ vertices encodes the game rock-paper-scissors. The one on $5$ vertices encodes the game rock-paper-scissors-lizard-Spock (if you don't know it, you can figure it out once you know that "lizard poisons Spock" [edit, thanks to Ramiro de la Vega:] and that "paper disproves Spock"). From $7$ on there is some choice, how much and which?

Also: does someone know the proper expression for "balanced"? [edit: according to David Speyer the term in this context is "regular tournament"]