Let $n$ be a positive integer. A set of $n$ distinct lines divides the plane into various (possibly unbounded) regions. The set of lines is called "nice" if no three lines intersect at a single point. A "colouring" is an assignment of two colours to each region such that the first colour is from the set $\left\{A_{1}, A_{2}\right\}$, and the second colour is from the set $\left\{B_{1}, B_{2}, B_{3}\right\}$. Given a nice set of lines, we call it "colourable" if there exists a colouring such that

1). no colour is assigned to two regions that share an edge;
2). for each $i \in\{1,2\}$ and $j \in\{1,2,3\}$ there is at least one region that is assigned with both $A_{i}$ and $B_{j}$.

Determine all $n$ such that every nice configuration of $n$ lines is colourable.
The answer is $n \geq 5$. If $n \leq 4$, consider $n$ parallel lines. There are 6 total colour combinations required, and only $n+1 \leq 5$ total regions, hence the colouring is not possible.

Now, assume $n \geq 5$. Rotate the picture so that no line is horizontal, and orient each line so that the "forward" direction increases the $y$-value. In this way, each line divides the plane into a right and left hand side (with respect to this forward direction). Every region of the plane is on the right hand side of $k$ lines and on the left hand side of $n-k$ lines for some $0 \leq k \leq n$. Furthermore, there is a region for every $k$ : let $w$ be large enough so that $w$ is greater than the $y$-value of any intersection point of two lines. Consider the horizontal line $y=w$ : a point very far on the left of this line is left of every single line, and as we cross over all lines in the problem, we hit all values of $k$.

Finally, take a region that is on the right hand side of $k$ lines. Colour it $A_{1}$ if $k$ is odd, and $A_{2}$ if it is even. Similarly, colour it $B_{i}$ if $k \equiv i(\bmod 3)$. By the previous paragraph, there are regions for at least $k=0,1, \ldots, 5$, whence there is a region coloured $A_{i}$ and $B_{j}$ for all $(i, j)$. Furthermore, two regions that share an edge will be on the right hand side of $k$ and $k+1$ lines for some $k$. By construction, the $A_{i}$ and $B_{i}$ colours of the regions must differ, hence we have proven that the set of lines is colourable.