3. Can 100 lines in a plane have exactly 2005 intersection points?
3. 100 lines intersecting pairwise can yield $\mathrm{C}_{100}^{2}=4950$ intersection points, which is 2945 more than 2005.

If $k$ lines are adjusted to be parallel, it can reduce $\mathrm{C}_{k}^{2}$ intersection points. If $k$ lines are adjusted to be concurrent, it can reduce $\mathrm{C}_{k}^{2}-1$ intersection points.

Since $\mathrm{C}_{77}^{2}=2926, \mathrm{C}_{6}^{2}=15, 2\left(\mathrm{C}_{3}^{2}-1\right)=4, 2926+15+4=2945$. Therefore, take one group of 77 parallel lines, another group of 6 parallel lines, and two groups of three concurrent lines, with the remaining 11 lines intersecting all lines without any three lines being concurrent. Thus, 100 lines have exactly 2005 intersection points.