Solution:

The six even chairs form a cycle (chair $12$ is adjacent to chair $1$), and each odd chair lies between two consecutive even chairs. The ambassadors occupy $4$ of the $6$ even chairs, and each advisor must take one of the two odd chairs flanking their ambassador, with all choices distinct. For a maximal block of $k$ consecutive occupied even chairs, the $k$ occupants choose among the $k + 1$ odd chairs touching the block; recording each choice as left or right, a conflict occurs exactly when someone picks right and their neighbor picks left, so the valid patterns are the strings of $L$s followed by $R$s: $k + 1$ of them.

Now case on the two empty even chairs among the six positions. If they are adjacent ($6$ ways), the occupied chairs form one block of $4,$ giving $5$ patterns. If they are separated by one chair ($6$ ways), the blocks have sizes $1$ and $3,$ giving $2 \cdot 4 = 8$ patterns. If they are opposite ($3$ ways), the blocks have sizes $2$ and $2,$ giving $3 \cdot 3 = 9$ patterns. The number of seat configurations is $6 \cdot 5 + 6 \cdot 8 + 3 \cdot 9 = 105.$

Finally, the four ambassador-advisor pairs can be assigned to the four chosen even chairs in $4! = 24$ ways, so $N = 105 \cdot 24 = 2520,$ and the remainder modulo $1000$ is $520.$