jmc

If we choose $a_3$ and $a_1$ such that $(10^3)(a_3)+(10)(a_1)\le 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3=0$ or $a_3=1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3=2$ then $a_1=0$ or $a_1=1$. Thus $N=100+100+2=\boxed{202}$.