jmc

Use complementary probability and Principle of Inclusion-Exclusion. If we consider the delegates from each country to be indistinguishable and number the chairs, we have$$\frac{9!}{(3!{)}^3}=\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{6\cdot 6}=6\cdot 8\cdot 7\cdot 5=30\cdot 56$$total ways to seat the candidates. Of these, there are $3\times 9\times \frac{6!}{(3!{)}^2}$ ways to have the candidates of at least some one country sit together. This comes to$$\frac{27\cdot 6\cdot 5\cdot 4}{6}=27\cdot 20.$$ Among these there are $3\times 9\times 4$ ways for candidates from two countries to each sit together. This comes to $27\cdot 4.$ Finally, there are $9\times 2=18.$ ways for the candidates from all the countries to sit in three blocks (9 clockwise arrangements, and 9 counter-clockwise arrangements). So, by PIE, the total count of unwanted arrangements is $27\cdot 20-27\cdot 4+18=16\cdot 27+18=18\cdot 25.$ So the fraction$$\frac{m}{n}=\frac{30\cdot 56-18\cdot 25}{30\cdot 56}=\frac{56-15}{56}=\frac{41}{56}.$$Thus $m+n=56+41=\boxed{097}.$