For an integer $n \geq 2$, consider the line segment connecting the point $(0, k)$ to the point $(n-k, 0)$ for $k=0,1,2, \ldots, n$. The union of these $n+1$ line segments divides the plane into one unbounded region and a number of bounded regions, each of which is a triangle or a quadrilateral. Each of these bounded regions can be coloured blue or red in a unique way such that regions sharing an edge have different colours and the region with vertex $(0,0)$ is coloured blue.

Determine all values of $n$ for which the total area that is coloured blue is equal to the total area that is coloured red.
For $k=0,1,2, \ldots, n$, let $\ell_{k}$ denote the line connecting the point $(0, k)$ to the point $(n-k, 0)$. For $0 \leq i \leq j \leq n$, define the point
$$
P_{i j}=\left(\frac{(n-i)(n-j)}{n}, \frac{i j}{n}\right) .
$$

One can check that for $i<j$, the point $P_{i j}$ is the point where the lines $\ell_{i}$ and $\ell_{j}$ intersect. Furthermore, observe that for $j=0,1,2, \ldots, n$, the $n+1$ points
$$
P_{0 j}, P_{1 j}, P_{2 j}, \ldots, P_{j-1, j}, P_{j j}, P_{j, j+1}, \ldots, P_{j n}
$$
are equally spaced along $\ell_{j}$ in order, with $P_{0 j}$ and $P_{j n}$ forming the endpoints.
We will prove that the area that is coloured blue is equal to the area that is coloured red if and only if $n$ is odd.

Let $n$ be odd and consider the part of the diagram below the line $y=x$. By drawing the blue segments shown below in the case of $n=7$, we split this part of the diagram into triangles, each of which pairs with an adjacent triangle with the same area but different colour. The fact that the areas are equal follows from the fact above concerning the equidistant spacing of the points along each line. It follows that the area that is coloured blue is equal to the area that is coloured red.
Now let $n$ be even and consider the part of the diagram below the line $y=x$. By drawing the blue segments shown below in the case of $n=6$, we split this part of the diagram into triangles, most of which which pair with an adjacent triangle with the same area but different colour. The fact that the areas are equal follows from the fact above concerning the equidistant spacing of the points along each line. However, in this case, the "edge" triangles are blue and are twice the area of the triangle that they pair with. It follows that the area that is coloured blue is greater than the area that is coloured red.