Note that $$\begin{aligned} 2 \cdot 6 \cdot 10 \cdots(4 n-2) & =2^{n} \cdot 1 \cdot 3 \cdot 5 \cdots(2 n-1) \\ & =2^{n} \cdot \frac{1 \cdot 2 \cdot 3 \cdots 2 n}{2 \cdot 4 \cdot 6 \cdots 2 n} \\ & =\frac{1 \cdot 2 \cdot 3 \cdots 2 n}{1 \cdot 2 \cdot 3 \cdots n} \end{aligned}$$ that is, it is just $(2 n)!/ n$ !. Therefore, since $(2 n)!/(n!)^{2}=\binom{2 n}{n}$ is always an integer, the answer is 2005.