Intersection of n lines 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)
My guess is that, we cannot do it iff when n is even. 


You could be right! But we'll need a proof ...


  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


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!
In general, any $n$ random lines in the plane will be "almost surely" not crossing them in a common points with three lines. 

This is because, $n$ lines have $2n$ variables (i.e. degrees of freedom, DOF), and having a common point shared with three lines imposes a constraint over 1 degree of freedom (for fixing the slope or crossing of just the third line), leaving you with $2n-1$ variables. Hence, the space of solutions of $2n-1$ DOFs  in a space of $2n$ DOFs solution. This shows that the required case is not "so common". 

Hence, a lot of simple counterexample can be found trivially. For example if we take the tangents to the segment of the unit circle:


Take a unit circle,
Divide the hemi-circle $(0,\frac\pi2]$ in $\frac{\pi}{2n}$ angle segments, $n$ points,
Draw a tangent $\mathscr{L}_i$ at every division point $i=1...n$,
The lines $\mathscr{L}_i$ are the requested lines. 


Proof:


The lines are:
$$
\mathscr{L}_i:\frac{y-\sin(\theta_i)}{x-\cos(\theta_i)}=\tan(\theta_i+\frac\pi2)\\
y=-\frac{x}{\tan(\theta_i)}+\frac{1}{\sin^2(\theta_i)}
$$
which are all with different slope for any $i$ in $(0,\frac\pi2]$.
The crossing are given by:
$$
y_{ij}=-\frac{x_{ij}}{\tan(\theta_i)}+\frac{1}{\sin^2(\theta_i)}=-\frac{x_{ij}}{\tan(\theta_j)}+\frac{1}{\sin^2(\theta_j)}
$$
which are all different for any given pair of $i,j$.
$$
\blacksquare
$$