Suppose $G$ is the directed graph representing wins and losses in a valid tournament. Arrange its vertices in a circle to satisfy the given condition. Claim: For players $A$, $B$, and $C$ with $A$ and $B$ adjacent, exactly one of $A$ and $B$ won against $C$. Proof: This is a simple averaging argument. The expected number of edges leaving a randomly chosen pair of adjacent players $A$ and $B$ is $n-2$, so the expected number of players beaten by both $A$ and $B$ is 0. Now the structure of $G$ is completely determined. The edges leaving any player $C$ must alternately leave and enter $C$ as we go around the circle. It is not hard to check that only odd $n$ work.

My solution is a bit different. Since we will work with directed graphs, I will use the notation $(u\to v)$ to denote an edge from $u$ to $v$ (which also means that $v$ lost to $u$). To begin with, observe that any odd $n$ works. Let $n=2k+1$. Fix an arbitrary vertex $v$ and number the remaining $2k$ vertices clockwise, starting from the right neighbour of $v: v,v_1,...,v_{2k}$. Now for each $i\in\{1,2,...,k\}$ trace $(v\to v_{2i})$ and $(v_{2i-1}\to v)$. Then, proceed to do this for every initially chosen vertex $v$. It is easy to see that this configuration satisfies the statement's condition. We will now prove that any even $n$ does not work. Let $n=2k$. Just as before, fix an arbitrary vertex $v$ and number the remaining $2k-1$ vertices clockwise, starting from the right neighbour of $v: v,v_1,...,v_{2k-1}$. Observe that at least one of $(v_1,v_2), \ (v_3,v_4), \ ...$ and one of $(v_{2k-3},v_{2k-2})$ beat $v$. Therefore, $\deg^{-}(v)\geq k-1$. Now, lets take a look at $v_1$. Assume $v_j$ is a vertex that beats $v$, different from $v_1$. According to our condition, since at least one of $v$ and $v_1$ must beat $v_j$ then we must have $(v_1\to v_j)$. Therefore, if $v_j$ ($j\neq 1$)beats $v$, then $v_1$ must beat $v_j$. It is trivial to see that this implies $\deg^{+}(v_1)\geq \deg^{-}(v)$ and thus, $\deg^{+}(v_1)\geq k-1$. However, we are dealing with a complete graph, so $\deg^{-}(v_1)+\deg^{+}(v_1)=2k-1$ and because $\deg^{+}(v_1)\geq k-1$ then $\deg^{-}(v_1)\leq k$. But remember that $v$ was chosen arbitrarily, so $v_1$ is also "chosen" arbitrarily. Therefore, for any vertex $u$, we know that: \[k\geq \deg^{-}(u)\geq k-1\text{ and that if }w\text{ is the vertex to the right of }u\text{ then }\deg^{+}(w)\geq \deg^{-}(u).\] From the first condition, one can easily see that a vertex $v$ we must either have: \[\big\{\deg^{+}(v)=k, \ \deg^{-}(v)=k-1\big\} \ \text{(let this be a $t_1$ i.e. a type $1$ vertex) or }\big\{\deg^{+}(v)=k-1, \ \deg^{-}(v)=k\big\} \ \text{(let this be a $t_2$ i.e. a type $2$ vertex)}.\] From the second condition, we can deduce that to the right of a $t_1$ can be either a $t_1$ or a $t_2$, but to the right of a $t_2$ there can only be a $t_2$. Let's see what this means. Assume our arbitrary $v$ is a $t_2$. Well, because to the right of a $t_2$ must be a $t_2$, then $v_1$ is also a $t_2$ so $v_2$ is also a $t_2$ and so on... we get the fact that all of the vertices are $t_2$s. Assume our arbitrary $v$ is a $t_1$. Then $v_{2k-1}$ is a $t_1$ too, becaue $v$ is to the right of $v_{2k-1}$ and to the right of a $t_2$ there can only be a $t_2$. Therefore, $v_{2k-1}$ is a $t_1$ so $v_{2k-1}$ is a $t_1$ too and so on... and we get the fact that all the vertices are $t_1$s. Therefore, in either case we get the fact that for any vertices $u$ and $v$ we have $\deg^{-}(u)=\deg^{-}(v)=c$ for some constant $c$. however, it is well known that: \[\sum_{v\in G}\deg^{-}(v)=|E(G)| \ \text{(Here, $G$ is an oriented graph and $E(G)$ is the set of edges of $G$)}.\]Applying this in our case, we get \[n\cdot c=\binom{n}{2}=\frac{n(n-1)}{2}\iff 2k\cdot c=k(2k-1)\]which is an obvious contradiction. Therefore, the only integers $n$ that satisfy the statement are the odd integers.