Answer

$\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=10 \Leftrightarrow a^{2}b-10ab^{2}+a-10b=0 \Leftrightarrow (ab+1)(a-10b)=0$ $\therefore$ either $ab+1=0$ or $a-10b=0$. The former is impossible as $a,b>0$, the latter means $a=10b$. $a+b\leq 100 \Rightarrow 11b\leq 100 \Rightarrow b\leq 9$ as $b$ is an integer. So solutions for $(a,b)$ are $(10,1);(20,2);(30,3);(40,4);(50,5);(60,6);(70,7);(80,8);(90,9)$ so there are indeed 9 solutions.

Since \[ \frac{a+\frac1b}{\frac1a+b} = \frac{\frac{ab+1}{b}}{\frac{ab+1}{a}} = \frac ab \]we seek the number of solutions to $a=10b$ with $a+b\le100$, which is nine.

Exactly: 10b+b<=100 which gives b<=9 And hence we have 9 solutions