This is too easy and I know it....

$\begin{aligned} S & = \frac 1{\sqrt 1+\sqrt 2} + \frac 1{\sqrt 2+\sqrt 3}+ \frac 1{\sqrt 3+\sqrt 4} + \cdots + \frac 1{\sqrt{98} + \sqrt{99}} + \frac 1{\sqrt{99} + \sqrt{100}} \\ & = \small \frac {\sqrt 2-\sqrt 1}{(\sqrt 2+\sqrt 1)(\sqrt 2-\sqrt 1)} + \frac {\sqrt 3-\sqrt 2}{(\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)} + \frac {\sqrt 4-\sqrt 3}{(\sqrt 4+\sqrt 3)(\sqrt 4-\sqrt 3)} + \cdots + \frac {\sqrt {99}-\sqrt {98}}{(\sqrt {99}+\sqrt {98})(\sqrt {99}-\sqrt {98})} + \frac {\sqrt {100}-\sqrt {99}}{(\sqrt {100}+\sqrt {99})(\sqrt {100}-\sqrt {99})} \\ & = \frac {\sqrt 2-\sqrt 1}{2-1} + \frac {\sqrt 3-\sqrt 2}{3-2} + \frac {\sqrt 4-\sqrt 3}{4-3} + \cdots + \frac {\sqrt {99}-\sqrt {98}}{99-98} + \frac {\sqrt {100}-\sqrt {99}}{100-99} \\ & = \sqrt 2-\sqrt 1 + \sqrt 3-\sqrt 2+\sqrt 4-\sqrt 3 + \cdots + \sqrt {99}-\sqrt {98} + \sqrt {100}-\sqrt {99} \\ & = \cancel{{\color{#3D99F6}\sqrt 2}} -\sqrt 1 + \cancel{{\color{#D61F06}\sqrt 3}}-\cancel{{\color{#3D99F6}\sqrt 2}} + \cancel{{\color{#3D99F6}\sqrt 4}}- \cancel {{\color{#D61F06} \sqrt 3}} + \cdots + \cancel{\color{#D61F06}\sqrt {99}}-\cancel{\color{#3D99F6}\sqrt {98}} + \sqrt {100}-\cancel{\color{#D61F06}\sqrt {99}} \\ & = \sqrt{100} - \sqrt 1 = 10 - 1 = \boxed 9 \end{aligned}$

We have:

$\Sigma_{k=1}^{99} \frac{1}{\sqrt{k+1} + \sqrt{k}} = \Sigma_{k=1}^{99} \frac{1}{\sqrt{k+1} + \sqrt{k}} \cdot \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}}$ ;

or $\Sigma_{k=1}^{99} \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k}$ ;

or $\Sigma_{k=1}^{99} \sqrt{k+1} - \sqrt{k}$ .

This is of course a telescoping series, which sums to $\sqrt{100} - \sqrt{1} = \boxed{9}.$