#22 Hua Luo Geng.

Relevant wiki: Partial fractions , Telescoping series

$\begin{aligned} S&=\sum_{n=1}^{100}\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}}\\ &=\sum_{n=1}^{100}\sqrt{\dfrac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}}\\ &=\sum_{n=1}^{100}\sqrt{\dfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2}}\\ &=\sum_{n=1}^{100}\sqrt{\dfrac{n^4+n^2+1+2n^3+2n^2+2n}{n^2(n+1)^2}}\\ &=\sum_{n=1}^{100}\sqrt{\dfrac{(n^2+n+1)^2}{n^2(n+1)^2}}\\ &=\sum_{n=1}^{100}\dfrac{n^2+n+1}{n(n+1)}\\ &=\sum_{n=1}^{100}\dfrac{n(n+1)+1}{n(n+1)}\\ &=\sum_{n=1}^{100}\mathbin{\color{#3D99F6}1}+\mathbin{\color{#D61F06}\dfrac{1}{n(n+1)}}\\ &=\mathbin{\color{#3D99F6}100}+\sum_{n=1}^{100}\mathbin{\color{#D61F06}\dfrac{1}{n}-\dfrac{1}{n+1}}\phantom{ccccccccicccc}\color{#D61F06}\text{by partial fractions}\\ &=100+\mathbin{\color{#D61F06}1-\dfrac{1}{101}}\phantom{ccccccccccccccccccc}\color{#D61F06}\text{by telescoping series sum}\\ &=101-\dfrac{1}{101}=a-\dfrac{1}{b} \end{aligned}$

$\Longrightarrow a+b=101+101=\boxed{202}$