Answer: 521

Small Hint:

Factor $n^2 - 4 = (n-2)(n+2)$ and use partial fractions to make the sum telescope

Big Hint:

Only $1,$ $\frac{1}{2},$ $\frac{1}{3},$ $\frac{1}{4}$ survive at the front, minus four tiny tail terms near $\frac{1}{10000}$

Solution:

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

Let $r$ be $250$ times the sum of the four positive tail fractions. Then $0 \lt r \lt 250 \cdot \frac{4}{9999} \lt \frac{1}{3}.$ The value of the expression is $520 + \frac{5}{6} - r,$ so it lies strictly between $520 + \frac{1}{2}$ and $520 + \frac{5}{6}.$ Hence the closest integer is $521.$