$$
\begin{aligned}
& \frac{2}{\log _4 2000^6}+\frac{3}{\log _5 2000^6} \\
& =\frac{\log _4 16}{\log _4 2000^6}+\frac{\log _5 125}{\log _5 2000^6} \\
& =\frac{\log 16}{\log 2000^6}+\frac{\log 125}{\log 2000^6} \\
& =\frac{\log 2000}{\log 2000^6} \\
& =\frac{\log 2000}{6 \log 2000} \\
& =\frac{1}{6}
\end{aligned}
$$
Therefore, $m+n=1+6=007$
Solution 2
Alternatively, we could've noted that, because $\frac{1}{\log _a b}=\log _b a$
$$
\begin{aligned}
\frac{2}{\log _4 2000^6}+\frac{3}{\log _5 2000^6} & =2 \cdot \frac{1}{\log _4 2000^6}+3 \cdot \frac{1}{\log _5 2000^6} \\
& =2 \log _{2000^6} 4+3 \log _{2000^6} 5 \\
& =\log _{2000^6} 4^2+\log _{2000^6} 5^3 \\
& =\log _{2000^6} 4^2 \cdot 5^3 \\
& =\log _{2000^6} 2000 \\
& =\frac{1}{6}
\end{aligned}
$$
Therefore our answer is $1+6=7$.
