suppose that $x = 2k+1$. ($k \leq 1002$) $(x+1)(2x-1)^2 = 2y^2 \Longrightarrow (k+1)(4k+1)^2 = y^2$. so, if $(k+1)$ is a square, our equation has a solution. now, we easily see that we have 31 perfect squares $\leq 1002$

This was given at an Australian training in 2004. As well as the straightforward solution posted by e.lopes, there's a weird one which uses the fact that substituting $x=\cos \alpha$ and $y=\cos\beta$ into the equation yields $\cos 3\alpha=\cos 2\beta$. I've just tried to reconstruct it but I can't remember exactly how it goes.

I hope this is the solution Agrippina had in mind. Notice that $\cos (ix)=\frac{e^x+e^{-x}}{2}$. Thus from the previous post, it is sufficient to prove that there are at least $31$ values of $x$ such that $\frac{e^{2x}+e^{-2x}}{2}\le 2005$ and $b=\frac{e^{3x}+e^{-3x}}{2}$ and $a=\frac{e^{2x}+e^{-2x}}{2}$ are both integers. Let $c=\frac{e^x+e^{-x}}{2}\in \mathbb Z$. Thus, $a=2c^2-1\in \mathbb Z$ and $b=4c^3-3c\in \mathbb Z$. Notice that $1\le c\le 31 \le \sqrt{1003}$. Thus, there are at least $31$ such values of $x$, as desired.

oh yes!the key step is to notice $LHS=(x+1)(2x-1)^2$. besides,what's 'Donova Olympiad'?