jmc

Note that $$\frac{1}{n^2-4}=\frac{1}{(n-2)(n+2)}=\frac{1}{4}\left(\frac{1}{n-2}-\frac{1}{n+2}\right).$$Thus, the given sum telescopes: $$\begin{array}{rl}1000\sum _{n=3}^{10000}\frac{1}{n^2-4}& =1000\cdot \frac{1}{4}\sum _{n=3}^{10000}\left(\frac{1}{n-2}-\frac{1}{n+2}\right)\\ & =250\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9999}-\frac{1}{10000}-\frac{1}{10001}-\frac{1}{10002}\right)\\ & =250+125+83.\overline{3}+62.5-\epsilon \end{array}$$where $\epsilon =250\left(\frac{1}{9999}+\frac{1}{10000}+\frac{1}{10001}+\frac{1}{10002}\right)$. This simplifies to $520.8\overline{3}-\epsilon$, and so the answer is $\boxed{521}.$

(To check that $\epsilon$ is small enough to not affect the answer, we can write $\epsilon <250\cdot 4\cdot \frac{1}{5000}=0.2$. This shows that the sum lies between $520.8\overline{3}$ and $520.6\overline{3}$, and so the closest integer is indeed $521$, as stated before.)