Using
$$\frac{1}{\sqrt{n} + \sqrt{n+2}} \le \frac{1}{4} \left( \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+2}} \right)$$
we obtain
$$\begin{align*} S &= \frac{1}{1+\sqrt{3}} + \frac{1}{\sqrt{5}+\sqrt{7}} + \cdots + \frac{1}{\sqrt{97}+\sqrt{99}} \\ < & \frac{1}{4} \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} + \cdots + \frac{1}{\sqrt{99}} \right) \\ &= \frac{1}{4} + \frac{1}{4} \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} + \cdots + \frac{1}{\sqrt{99}} \right) \\ < & \frac{1}{4} + \frac{1}{2} \left( \frac{1}{\sqrt{1}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{5}} + \cdots + \frac{1}{\sqrt{97}+\sqrt{99}} \right) \\ &= \frac{1}{4} + \frac{1}{4} \left( \sqrt{3} - \sqrt{1} + \sqrt{5} - \sqrt{3} + \cdots + \sqrt{99} - \sqrt{97} \right) \\ &= \frac{1}{4} + \frac{1}{4} \left( \sqrt{99} - 1 \right) \\ &= \frac{\sqrt{99}}{4} = 2 + \epsilon, \quad 0 < \epsilon < 1. \end{align*}$$
On the other hand,
$$\begin{align*} S &> \frac{1}{2} \left( \frac{1}{\sqrt{1}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{5}} + \frac{1}{\sqrt{7}+\sqrt{9}} + \cdots + \frac{1}{\sqrt{99}+\sqrt{101}} \right) \\ &= \frac{1}{4} \left( \sqrt{3} - \sqrt{1} + \sqrt{5} - \sqrt{3} + \cdots + \sqrt{101} - \sqrt{99} \right) \\ &= \frac{1}{4} (\sqrt{101} - 1) = 2 + \delta, \quad 0 < \delta < 1. \end{align*}$$
Therefore the integer part of $S$ is $2$.