8. Given a regular 2000-gon $A_{1} A_{2} \cdots A_{2000}$, consider 2000 diagonals $A_{i} A_{i+17}(i=1,2, \cdots, 2000)$, where $A_{i+2000}=A_{i}(i=1,2, \cdots, 17)$. Then the number of different intersection points of these 2000 diagonals inside the regular 2000-gon is $\qquad$
8. 32000 .

For each $i=1,2, \cdots, 2000, A_{i+1}, A_{i+2}$, $\cdots, A_{i+16}$ each draw two diagonals intersecting with $A_{i} A_{i+17}$. Since each such intersection point is counted twice, the number of distinct intersection points of these 2000 diagonals inside the regular 2000-gon does not exceed $\frac{2 \times 16 \times 2000}{2}=32000$.

Since each of these diagonals is of equal length, these 2000 diagonals are all tangent to a fixed circle. Because only two tangents can be drawn from an intersection point, it is impossible for three such diagonals to intersect at one point. Therefore, these 2000 diagonals have 32000 distinct intersection points inside the regular 2000-gon.