Answer: 521

Solution:

Since $\frac{1}{n^2 - 4} = \frac{1}{4}\left(\frac{1}{n-2} - \frac{1}{n+2}\right),$ the sum telescopes: $$\begin{aligned} &1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4} \\ &\tiny{}= 250\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}\right). \end{aligned}$$

The front part is $250 \cdot \frac{25}{12} = 520.8\overline{3},$ and the four tail terms subtract only about $250 \cdot \frac{4}{10000} = 0.1.$ The value is therefore about $520.73,$ so the closest integer is $521.$