The answer is 139.

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We first count the total numer $T$ of combinations

$T=\binom{9}{3}=84$

And the number $F$ of favorable cases, counting the number of ways we can choose $2$ delegates from a country and multiplying it by the number of ways we can choose the third one from another country

Mexico: $F_M=\binom{2}{2}\cdot(9-2)=7$

Canada: $F_C=\binom{3}{2}\cdot(9-3)=18$

United States: $F_{US}=\binom{4}{2}\cdot(9-4)=30$

$\Longrightarrow F=7+18+30=55$

Therefore the probability is

$P=\dfrac{F}{T}=\dfrac{55}{84}=\dfrac{m}{n}$

$\Longrightarrow m+n=55+84=\boxed{139}$