8. Select five points $X_{1}, X_{2}, \cdots, X_{5}$ on a plane, and connect each pair of these points with a blue straight line. Assume that no two blue lines are either perpendicular or parallel. For each point $X_{i}(1 \leqslant i \leqslant 5)$, for every blue line $l$ that does not pass through $X_{i}$, there is a red line passing through $X_{i}$ that is perpendicular to $l$. How many intersection points can all the red lines have at most?
8. 315 .

For each point $X_{i}$, 6 blue lines that do not pass through it can be drawn, so, 30 red lines can be drawn in total, which have $\mathrm{C}_{30}^{2}=435$ "intersection points" (these intersection points include the intersections of parallel lines and the cases where three or more lines intersect at one point, which are counted repeatedly).
Next, we calculate the number of these repeated points.
For each point $X_{i}$, 6 red lines passing through it can be drawn, and these lines have $\mathrm{C}_{6}^{2}=15$ intersection points coinciding, at this time, there are $5 \times(15-1)=70$ intersection points;

For $\mathrm{C}_{5}^{2}=10$ blue lines, each line will have three red lines perpendicular to it, thus, these three red lines are parallel and cannot intersect, at this time, there are $10 \mathrm{C}_{3}^{2}=30$ intersection points;

The three altitudes of any triangle will intersect at one point, and 5 points can form $\mathrm{C}_{5}^{3}=10$ triangles, so, at this time, there are $10 \times 2=20$ intersection points.
Therefore, the number of intersection points of all red lines is at most $435-70-30-20=315$.