$$\boxed{272020}$$
Let $a$ be the number of the sides of the 2010-gon lying in between the vertices $B$ and $C$ and on the opposite side from the vertex $A$, and $b$ be the number of the sides lying in between the vertices $C$ and $A$ and on the opposite side from the vertex $B$. Let $\Gamma$ be the circum-circle of this regular 2010-gon, and let $O$ be its center. Then, we have
$$\angle CAB = \frac{1}{2} \angle COB = \frac{1}{2} \times 360^\circ \times \frac{a}{2010} = \left( \frac{6a}{67} \right)^\circ$$
In the same way, we get $\angle ABC = \left(\frac{6b}{67}\right)^\circ$. When these two angles take the integral values, then the $\angle BCA$ also takes an integral value. Hence the condition of the problem is satisfied if and only if both $6a$ and $6b$ are divisible by $67$, which is the same as the statement that both $a$ and $b$ are multiples of $67$ (since $67$ is a prime).
From $\frac{2010}{67} = 30$, it also follows that this is equivalent to the statement that $A$, $B$, $C$ are the vertices of a regular 30-gon inscribed in the circle $\Gamma$. There are $\frac{2010}{30} = 67$ regular 30-gons whose vertices are chosen from the given regular 2010-gon and for each of these regular 30-gons there are $\binom{30}{3}$ ways of choosing 3 vertices. Therefore, there are $67 \times \binom{30}{3} = 272020$ triangles satisfying the condition of the problem.