We will prove that $n$ can be any odd number not less than $3$.

Suppose $n = 2k + 1$, an odd number greater than $3$, and $n$ players are represented by $A_1, A_2, \dots, A_{2k+1}$. Let us arrange the competition result as follows: $A_i$ ($1 \le i \le 2k+1$) defeated $A_{i+2}, A_{i+4}, \dots, A_{i+2k}$ (we stipulate that $A_{2k+1+j} = A_j$, $j = 1, 2, \dots, 2k+1$) but lost to the other players. Then these players can be arranged in a circle in order $A_1, A_2, \dots, A_{2k+1}, A_1$.

Now, given any three players $A$, $B$, $C$ with $A$, $B$ being adjacent in the circle, we may assume that $A = A_i$, $B = A_{i+1}$, $C = A_{i+r}$ ($1 \le i \le 2k+1$, $2 \le r \le 2k$). Then either $r$ or $r-1$ is an even number not less than $2k$, which implies that at least one of the players $A$, $B$ defeated $C$.

On the other hand, suppose $n$ is an even number not less than $4$, and the $n$ players can be arranged in a circle $A_1, A_2, \dots, A_n, A_1$ that meets the required condition. We may assume that $A_1$ defeated $A_2$. According to the requirement, at least one of $A_2$, $A_3$ defeated $A_1$, and then $A_3$ defeated $A_1$; but at least one of $A_1$, $A_2$ defeated $A_3$, so $A_2$ defeated $A_3$, and so forth. We then get that, for any $1 \le i \le n$, $A_i$ defeated $A_{i+1}$ and lost to $A_{i-1}$ (stipulate that $A_{n+1} = A_1$, $A_0 = A_n$).

We now divide the players after $A_i$ and before $A_{i-1}$ into $\frac{n-2}{2}$ pairs — each consists of two adjacent players. Then there is at least one player in each pair who defeated $A_i$, and that means, besides $A_{i-1}$, there are at least $\frac{n-2}{2}$ players who defeated $A_i$. Then $A_i$ lost at least $\frac{n}{2}$ games. So $n$ players lost totally at least $\frac{n^2}{2}$ games.

But the number of total games is $C_n^2 = \frac{n(n-1)}{2} < \frac{n^2}{2}$. This is a contradiction. Therefore, the possible values of $n$ are all the odd numbers not less than $3$.

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