Let $n$ lines in the plane be given such that no two of them are parallel and no three of them have a common point. We want to choose the direction on every line so that the following holds: if we go along any line in its direction and put numbers from 1 to $n-1$ on the intersection points then no two equal numbers appear at the same point. For which numbers $n$ is it possible?
My guess is that, we cannot do it iff when $n$ is even. For the case $n$ is even, I think we should find a point $p$ of intersection of two lines which lies in the middle of two lines (for each of them half of the intersection points in one side of p and the others on the other side of p)
You could be right! But we'll need a proof ...
No, that won't work as a proof, since it is false. Here is a counterexample:
The midpoint for line 1 is where it crosses with line 6
The midpoint for line 2 is where it crosses with line 6
The midpoint for line 3 is where it crosses with line 6
The midpoint for line 4 is where it crosses with line 9
The midpoint for line 5 is where it crosses with line 2
The midpoint for line 6 is where it crosses with line 4
The midpoint for line 7 is where it crosses with line 3
The midpoint for line 8 is where it crosses with line 3
The midpoint for line 9 is where it crosses with line 3
The midpoint for line 10 is where it crosses with line 6
Or, as a graph, where an arrow from $i$ to $j$ means that the midpoint of line $i$ is where it crosses with line $j$:
So: you see that no two lines share the same midpoint!