1.86 In a convex 1000-gon, 500 points are given. Together with the vertices of the polygon, there are a total of 1500 points, and no three points are collinear. The convex 1000-gon is divided into some triangles, with all 1500 points being the vertices of these triangles, and these triangles have no other vertices. Under such a division, how many triangles is the convex 1000-gon divided into?
[Solution] Let this polygon be divided into $n$ triangles, then the sum of the interior angles of these triangles is $n \cdot 180^{\circ}$. On the other hand, the contribution of the 1000 vertices of the polygon to the sum of the interior angles is the sum of all the interior angles of the polygon, which is $998 \times 180^{\circ}$. Each point inside the 1000-sided polygon contributes $360^{\circ}$ to the sum of the interior angles. Therefore, the sum of the interior angles of all the triangles should be
$$
998 \times 180^{\circ} + 500 \times 360^{\circ} = 1998 \times 180^{\circ} \text{. }
$$

Thus, it is known that the polygon is divided into $n=1998$ triangles.