Answer: 97

Solution:

Only the pattern of countries in the nine chairs matters, and all $\frac{9!}{3!\,3!\,3!} = 1680$ patterns are equally likely. The condition fails for some delegate exactly when both of his neighbors are compatriots, which happens exactly when some country's three delegates occupy three consecutive chairs. Let $A_i$ be the set of patterns in which country $i$'s delegates are consecutive.

There are $9$ triples of consecutive chairs, so $|A_i| = 9\binom{6}{3} = 180,$ choosing which $3$ of the remaining $6$ chairs go to one of the other countries. For two countries, after placing the first block ($9$ ways) the remaining six chairs form an arc containing $4$ triples of consecutive chairs, so $|A_i \cap A_j| = 9 \cdot 4 = 36.$ For all three, the circle must split into three consecutive triples ($3$ ways) assigned to the countries in $3!$ orders: $|A_1 \cap A_2 \cap A_3| = 18.$ By inclusion-exclusion, $$\begin{aligned} |A_1 \cup A_2 \cup A_3| &= 3 \cdot 180 - 3 \cdot 36 \\ &\quad {}+ 18 = 450. \end{aligned}$$

The probability is $1 - \frac{450}{1680} = 1 - \frac{15}{56} = \frac{41}{56},$ so $m + n = 41 + 56 = 97.$