16. On a plane, there are 100 lines. Can they contain 1985 different intersection points.
16. If 100 lines are neither pairwise parallel nor have three lines concurrent, then the number of intersection points is $C_{100}^{2}=4950$, which is clearly greater than 1985, and this state does not meet the problem's requirements. However, we can consider starting from this state and appropriately adjusting the positions of some lines to reduce the number of intersection points, such as adjusting the positions of some lines to make them concurrent and form a line bundle.

By translating some lines, we get $k$ line bundles, with the number of lines in each bundle being $n_{1}, n_{2}, \cdots, n_{k}\left(n_{i} \geqslant 3, i=1,2, \cdots, k\right)$. Since not all lines may have been adjusted, $n_{1}+n_{2}+\cdots+n_{k} \leqslant 100$. After adjustment, the number of intersection points in each bundle decreases by $C_{n_{i}}^{2}-1$ $(i=1,2, \cdots, k)$, thus the total number of intersection points decreases by $C_{n_{1}}^{2}-1+C_{n_{2}}^{2}-1+\cdots+C_{n_{k}}^{2}-1$.

To make the number of intersection points 1985 after adjustment, the total decrease in the number of intersection points should be $C_{n_{1}}^{2}-1+C_{n_{2}}^{2}-1+\cdots+C_{n_{k}}^{2}-1=C_{100}^{2}-1985=2965$.

Now, let's estimate $n_{1}: C_{77}^{2}=2925$, which is closest to 2965, so $n_{1}=77$. Therefore, we first translate 77 lines to form a line bundle, reducing the number of intersection points by 2925. This way, we only need to further adjust to reduce the number of intersection points by 40.

Estimate $n_{2}$: Since $C_{9}^{2}-1=35$ is closest to 40, $n_{2}=9$, meaning we then translate 9 of the remaining lines to form another line bundle, reducing the number of intersection points by 35. We then need to further adjust to reduce the number of intersection points by 5.

Estimate $n_{3}$: Since $C_{4}^{2}-1=5$, $n_{3}=4$, meaning we finally translate 4 of the remaining 14 lines to form another bundle, reducing the number of intersection points by 5. The remaining 10 lines remain unchanged.

After the above adjustments, the total number of intersection points decreases by 2965, meaning the 100 lines after adjustment have exactly 1985 different intersection points, solving the problem.