Denote by $V,E,F$ the number of vertices, edges, and faces of the polyhedron respectively. Using Euler's formula we have that $V-E+F=2$ or $V+F = 2024$. Note that each face corresponds to $\geq 3$ edges and each edge corresponds to $2$ faces, hence $3F\leq 2E$, or $F\leq 1348$, which implies $V\geq 2024 - 1348 = 676$. On the other hand, if we consider the graphs formed by the edges of each color, as they're connected, they must each have at least $V-1$ edges, so \[k(V-1)\leq E\Longrightarrow k\leq \frac{E}{V-1}\leq \frac{2022}{675}<3\]For $k=2$ consider the following construction: the vertex $A = (0,0,1)$ and circles $k_{1}, k_{2}, \dots, k_{337}$ which are the intersection of planes $z = \text{const}$ with the unit sphere. Construct an equilateral triangle in each circle $k_{i}$ and denote it as $B_{i}^{1}B_{i}^{2}B_{i}^{3}$ so that points $A, B_{1}^{1}, B_{2}^{1}, \dots , B_{337}^{1}$ lie on an arc on the unit sphere (and the same for $B_{i}^{2}$ and $B_{i}^{3}$ respectively). This is a non-rigorous sketch, but I hope the construction is clear. Take $P$ to be the the convex hull of the points $A$, $B_{i}^{1}$, $B_{i}^{2}$, $B_{i}^{3}$ for all $i$. Hence every vertex is of $\deg 3$ (points $A$, $B_{337}^{1}$, $B_{337}^{2}$, $B_{337}^{3}$) or $4$ (all other points), so the second condition is satisfied. Also, $P$ has $2022$ edges because there are $3\times 337$ equilateral triangles, $3\times 336$ edges between different triangles and $3$ edges from $A$. Notice that here the number of vertices is $1012$, so the two graphs must be trees. A possible coloring is the following: \[\text{Green Edges} = \{AB_{1}^{1}, B_{i}^{1}B_{i}^{2}, B_{i}^{2}B_{i}^{3}, B_{i}^{1}B_{i+1}^{1}\}\]\[\text{Red Edges} = \{AB_{1}^{2}, AB_{1}^{3}, B_{i}^{1}B_{i}^{3}, B_{i}^{2}B_{i+1}^{2}, B_{i}^{3}B_{i+1}^{3}\}\]Hence this construction for $k=2$ works and we proved $k<3$, so we're done.

Marinchoo wrote: The condition in the first bullet is artificial and kind of dumb. We proved $k<3$ for all polyhedrons without using it, so it's only purpose is to make the construction for $k=2$ harder to find. That's true, and they actually specify that in a comment. This problem isn't the best, but it has a nice theoretical background when it comes to the duality between polyhedron graphs and planar graphs. As suggested by the title, this is juts a convoluted graph problem after all. The proof for $k<3$ works for planar graphs flawlessly. To exhibit a construction for $2(n-1)$ instead of $2022$ we use the following theorem: Theorem (Steinitz). A planar graph is the graph of a convex polyhedron if and only if it is 3-vertex connected (i.e. by removing any two vertices, the graph remains connected). Take the path $v_1,v_2,\ldots,v_{2n}$ and connect the vertices $v_i,v_{i+2}$ for all $i=1,\ldots,2n$ except $i=2n-1$, of course, indices being reduced modulo $2n$. This graph has $2(n-1)$ edges and the degrees of the vertices are three or four. It can be easily checked to be 3-vertex connected, hence it is a valid example.

This problem was proposed by Viktor Simjanoski.