I sense a pattern!

$\dfrac{1}{2^{2}-1}+\dfrac{1}{4^{2}-1}+\dfrac{1}{6^{2}-1}+\dfrac{1}{8^{2}-1}+\ldots+\dfrac{1}{1000^{2}-1}$

$=\dfrac{1}{(2-1)(2+1)}+\dfrac{1}{(4-1)(4+1)}+\dfrac{1}{(6-1)(6+1)}+\dfrac{1}{(8-1)(8+1)}+\ldots+\dfrac{1}{(1000-1)(1000+1)}$

$=\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5}+\dfrac{1}{5 \times 7}+\dfrac{1}{7 \times 9}+\ldots+\dfrac{1}{999 \times 1001}$

Now by Telescoping series,

$\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5}+\dfrac{1}{5 \times 7}+\dfrac{1}{7 \times 9}+\ldots+\dfrac{1}{999 \times 1001}$

$=\dfrac{1}{2}(\dfrac {1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7} \ldots-\dfrac{1}{1001})$

$=\dfrac{1}{2}(1-\dfrac{1}{1001})$

$=\dfrac{1}{2}(\dfrac{1000}{1001})$

$=\dfrac{500}{1001}$

Therefore, the answer is $500+1001=\boxed{1501}$

Let the expression be $S$ , then we have:

$\begin{aligned} S & = \sum_{k=1}^{500} \frac{1}{(2k)^2 - 1} \\ & = \sum_{k=1}^{500} \frac{1}{(2k - 1)(2k+1)} \\ & = \sum_{k=1}^{500} \frac{1}{2} \left( \frac{1}{2k - 1} - \frac{1}{2k + 1} \right) \\ & = \frac{1}{2} \left( \sum_{k=1}^{500} \frac{1}{2k - 1} - \sum_{k=1}^{500} \frac{1}{2k + 1} \right) \\ & = \frac{1}{2} \left( \sum_{k=1}^{500} \frac{1}{2k - 1} - \sum_{k=2}^{501} \frac{1}{2k - 1} \right) \\ & = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{1001} \right) \\ & = \frac{500}{1001} \end{aligned}$

$\Rightarrow a +b = 500+1001 = \boxed{1501}$