Let's change the language. We call the lines vertices of a graph, and the labels of the intersection points colors of the edges between the vertices. The edges colored with a certain color form a perfect matching in our graph, so the number of vertices must be even (i.e. $n$ is even). The fact that if $n$ is even we can always do the labeling is a different wording for the fact that the chromatic index of $K_{2n}$ (the complete graph on $2n$ vertices) is $2n-1$. In order to find a proper coloring of the edges of $K_{2n}$, we first find one for $K_{2n-1}$: Call the vertices of $K_{2n-1}$ $v_0,\ldots,v_{2n-2}$, and color the edge $v_iv_j$ with color $i+j\pmod{2n-1}$. Each vertex $v_i,\ i\in\overline{0,2n-2}$ is missing a color $c_i$, and the colors $c_i$ are all different (this is where we use the fact that $2n-1$ is odd, which implies that all of the $2i\pmod{2n-1}$ are distinct). Now add the $2n$'th vertex $v_{2n-1}$, and color the edges $v_{2n-1}v_i$ with color $c_i$.