jmc

We have $$\begin{array}{rl}{\log }_a\frac{a}{b}+{\log }_b\frac{b}{a}=& {\log }_{a}a-{\log }_{a}b+{\log }_{b}b-{\log }_{b}a\\ =& 1-{\log }_{a}b+1-{\log }_{b}a\\ =& 2-{\log }_{a}b-{\log }_{b}a.\end{array}$$Let $c={\log }_{a}b$, and note that $c>0$ since $a$ and $b$ are both greater than 1. Thus $${\log }_a\frac{a}{b}+{\log }_b\frac{b}{a}=2-c-\frac{1}{c}=\frac{c^2-2c+1}{-c}=\frac{(c-1{)}^2}{-c}\le 0.$$This expression is 0 when $c=1$, that is, when $a=b$. Hence the answer is $\boxed{0}$.