Draw line segments from $P_1$ to the points $A, B, C, D, E, F$, we get six smaller triangles. Since any three points in $S$ are not collinear, $P_2$ is inside one of the six triangles. Draw line segments joining $P_2$ with the vertices of this triangle to divide it into three triangles. The total number of triangles is increased by $2$. Do the same for the points $P_3, P_4, \dots, P_{2556}$ to subdivide the hexagon into non-overlapping triangles, each time the total number of triangles is increased by $2$. Thus, after the construction, the total number of triangles is $6 + (2555 \times 2) = 5116$. If the area of each of these triangles is $\ge \frac{1}{1700}$, then the area of the hexagon is $\ge 5116 \times \frac{1}{1700} > 3 > \frac{3\sqrt{3}}{2} = \text{area of the regular hexagon } ABCDEF$, which is a contradiction.