Solution:

Answer: $n^{2}$.

Obviously there cannot be more than $n^{2}$ people. For if there were, then at least one country would have more than $n$ representatives. But there are only $n$ different countries to choose their right-hand neighbours from. Contradiction.

Represent someone from country $i$ by $i$. Then for $n = 2$, the arrangement $1122$ works. [It wraps round, so that the second $2$ is adjacent to the first $1$.] Suppose we have an arrangement for $n$. Then each of $11, 22, \ldots, nn$ must occur just once in the arrangement. Replace $11$ by $1(n+1)11$, $22$ by $2(n+1)22$, $\ldots$, and $(n-1)(n-1)$ by $(n-1)(n+1)(n-1)(n-1)$. Finally replace $nn$ by $n(n+1)(n+1)nn$. It is easy to check that we now have an arrangement for $n+1$. We have added one additional representative for each of the countries $1$ to $n$ and $n+1$ representatives for country $n+1$, so we have indeed got $(n+1)^{2}$ people in all. We have also given a representative of each country $1$ to $n$ a neighbour from country $n+1$ on his right and we have given the ($n+1$) representatives from country $n+1$ neighbours (on their right) from each of the other countries. Otherwise we have left the seating unchanged.