, 2010

All the interior angles of the $2010$-gon can be built from the inner angles of the triangular pieces. As the sum of the inner angles of the $2010$-gon is $2008 \cdot 180^{\circ}$ and that of every triangle is $180^{\circ}$, there must be at least $2008$ triangles. On the other hand, each convex $2010$-gon can be divided into exactly $2008$ triangles by choosing one vertex and cutting the figure into pieces along the diagonals that start from this vertex.