Since the transitively reduced graph $G$ is acyclic, there is a corresponding undirected simple graph $G'$, which can be formed by ignoring the direction of the edge. The number of edges in $G$ is the same as the number of edges in $G'$.
For $G$ to be transitively reduced, a necessary condition is that $G'$ must not contain any triangles.
It is well known (Turan's/Mantel's theorem) that any simple undirected graph on $n$ vertices and more than $n^2/4$ edges has a triangle.
It is also known that the undirected graph on $n$ vertices with maximum edges and no triangles is the complete bipartite graph $K_{[n/2],[n/2]}$ (for instance check out: exercise 4)
From your comments, it looks like you found a $G$ whose corresponding $G'$ is $K_{[n/2],[n/2]}$ and so that proves it.