I'll use friend and enemy instead of defense agreement. Suppose players are in $k$ rooms namely $a_1,a_2...,a_k$. Obviously $k\leq n$. we claim $n+1$ rooms are enough to grant everyone's wish. Suppose the player from Fire country is in room $a_j$. WLOG we may assume in every other room $a_i$ with $i<j$, he has at least one enemy. If not, we can simply move him to room $a_i$. We allot rooms to coaches in the following manner. Since players of room $a_1$ are friends of one another, so their coaches must be placed into separate rooms namely $b(1,1)...b(1,a_1)$. Now take any player from room $a_2$. He has an enemy in room $a_1$. So his coach can be placed into one of the previous rooms. We allot separate room to every other coach of the players of room $a_2$. Also number these rooms $b(2,1),...,b(2,a_2-1)$. Now take a player $A$ from room $a_i,i>2$. We'll place his coach in the first room he fits in, starting from the last room $b(i-1,a_{i-1}-1)$. We claim there is at least one such room. A has an enemy country player $B_j$ in room $a_{j},j<i$. If the coach of $B_{i-1}$ is in one of the rooms $b(i-1,1),...,b(i-1,a_{i-1}-1)$, we can simply place $A$'s coach in the same room. But if $B_{i-1}$'s coach is not in these rooms, it means he is in one of the previous of the rooms. Because of our special room allotting system, from room $a_{i-1}$, there is be only one player whose coach is in previous rooms. Hence, if $A$'s coach can't be placed in any of the rooms $b(i-1,1),...,b(i-1,a_{i-1}-1)$, then any coach of a player of room $a_{i-1}$ who is in any of the rooms $b(1,1),...,b(i-2,a_{i-2}-1)$, must be from an enemy country of $A$. Now inductively if $B_j$'s coach is in any of the rooms $b(i-j,1),...,b(i-j,a_{i-j}-1)$, we can place $A$'s coach with him.(Of course some of the $B_{i-1},...,B_{j+1}$'s coach might be with them, that is not a problem because they are all enemies.) But if not, it means he is in one of the previous of the rooms, namely $b(1,1),...,b(j-1,a_{j-1}-1)$. Because of our special room allotting system, from room $a_{j}$, there is be only one player whose coach is in previous rooms. Hence, if $A$'s coach can't be placed in any of the rooms $b(j,1),...,b(i-1,a_{i-1}-1)$, then any coach of a player of room $a_{j}$ who is in any of the rooms $b(1,1),...,b(j-1,a_{j-1}-1)$, must be from an enemy country of $A$. But $B_1$'s coach is among $b(1,1),...,b(1,a_1)$ and we can simply place $A$'s coach with him. We allot separate room to all other players coaches from room $a_i$. So the number of rooms needed for coaches $a_1+\sum_{k\ge i\ge 2}a_i-1=n-k+1$. So total number of rooms $k+n-k+1=n+1$ And if all the countries are friends, then $k=1$ and all the coaches have to be placed in separate rooms. So in that case, $n+1$ rooms are necessary. Hence $n+1$ is the minimum number of rooms.