$n$ can only be 25.

We first provide an example for the case $n = 25$. Consider 10 participants $A_1, A_2, \dots, A_{10}$ in group $\mathcal{A}$, and 15 participants $B_1, B_2, \dots, B_{15}$ in group $\mathcal{B}$. Suppose the games played between 2 participants in the same group end in a draw. For the games played between 2 participants in different groups, the results are given below, where $O$ means a draw, while an empty box means the group $\mathcal{B}$ participant wins.

The score of each participant in group $\mathcal{A}$ is $9+9=18$, where 9 points come from the games played between them. The score of each participant in group $\mathcal{B}$ is $14+(6+2\times4)=28$, where 14 points come from the games played between them. So the conditions are satisfied.

Now, note that the two participants in a game always gain 2 points in total. Therefore, the total score of all participants is
$$2\binom{n}{2} = n(n-1).$$

For the 10 participants with the lowest scores, their total score gained by the games played between them is $2\binom{10}{2} = 90$. Therefore, their total score is
$$2 \times 90 = 180.$$
For the other $n-10$ participants, their total score gained by the games played between them is $2\binom{n-10}{2} = (n-10)(n-11)$. Note that this is half of their total score since the remaining half comes from the other games. Therefore, their total score is
$$2(n - 10)(n - 11).$$

Combining all these, we need $180+2(n-10)(n-11) = n(n-1)$. This is the same as $(n-16)(n-25) = 0$. Thus, $n = 16, 25$.

For $n = 16$, the 10 participants with the lowest scores obtain 180 points in total, while the other 6 participants obtain $2(16 - 10)(16 - 11) = 60$ points in total. Since $\frac{180}{10} = 18 > 10 = \frac{60}{6}$, it is impossible that all those 10 participants have the lowest score. Therefore, $n = 16$ is rejected, and $n = 25$ is the only possibility.