Partial solution... specifically:
Claim A: Any odd $n$ is feasible.
Claim B: $n = 4$ is infeasible.
Proof of A: Arrange the players in a circle and number them $0, ..., n-1$, and let $i$ beat $i+1, i+3, ..., i+n-2$. All arithmetic is modulo $n$.
First of all, this assignment is consistent: For any $i \neq j$, if $j = i + odd$ (i.e. $i$ beats $j$) then $i = j + even$ (i.e. $j$ does not beat $i$).
Next, clearly $i$ beats all the $i+odd$ directly, but since each $j$ beats $j+1$, $i$ also indirectly beats all the $i+odd+1$, i.e. all the $i+even$.
Proof of B: Among the $n=4$ players, clearly nobody can beat everyone or be beaten by everyone. Since each plays $3$ games, that means each must win only $1$ or $2$ games. Since there are $6$ games total, the only way to do this is if two players $W,X$ win twice each and two other players $Y,Z$ win once each. But consider the match between $Y,Z$ and without loss assume $Y$ beats $Z$. This is $Y$'s only win, and $Z$ beats only $1$ person (e.g. $W$), so $Y$ does not directly nor indirectly beat the other person (e.g. $X$).