Solution:

Note that $G$ has no $K_{5}$; indeed, it's well-known that the only triangle-free coloring of the edges of $K_{5}$ consists of two monochromatic $5$-cycles. Therefore, the number of edges of $G$ is at most $\binom{4}{2} \cdot 505^{2} = 1530150$ by Turán's theorem.

To show this occurs, we split the graph into four equally sized components $A, B, C, D$. We color red all edges between $A$ and $B$, or $C$ and $D$. We color green all edges between $A$ and $C$, $A$ and $D$, $B$ and $C$, and $B$ and $D$. This indeed has the claimed number of edges, and the subgraphs formed by each color are bipartite, so this solves the problem.