Hmm, it's easy to show by Pigeonhole that $n^2+1$ doesn't work... I don't know how to easily explain a construction for $n^2$ though.

Indeed, the maximal number is : $n^2$. Like explained by paladin8, we know that our maximal number is less or equal than $n^2$. Now, let us denote the $n$ countries by $C_1, C_2, \ldots, C_n$. Also, for all $k$, we denote by $n_k$ the number of representants coming from $C_k$. So $N = \sum_{k=1}^n n_k$ is the total number of representants. And, $r_k$ denotes a representant of $C_k$. Ok, so, via induction on $n$, we can make the desired construction for $n^2$. For $n=2$, it's trivial to get a good construction. Now, let's take a fixed $n \ge 2$. Let us assume that $(a_1, a_2, \ldots, a_{n^2})$ is a good construction for $n$ countries. Then, for all $k$, we have $n_k \le n$. Moreover, since $\sum_{k=1}^n n_k = n^2$, it means that : $n_k = n$, $\forall k$. Since the construction is good, we have that each pair $(r_1, r_1), (r_2, r_2), \ldots, (r_n, r_n)$ appears exactly once in the construction. We then introduce the representants of the country $C_{n+1}$ in the following way : the couple $(r_1, r_1)$ is replaced with $(r_1, r_{n+1}, r_{n+1}, r_1, r_1)$. For $k \ge 2$, each couple $(r_k, r_k)$ is replaced with $(r_i, r_{n+1}, r_i, r_i)$. It is easy to check that our new construction is good, and, there are $n^2 + 3 + 2(n-1) = (n+1)^2$ representants, which proves the result via induction, and provides the desired result!