What a nice problem! The answer is $1009^2$, achieved by considering the graph of $K_{1010}$ and $K_{1009}$, and none of the vertex in the $K_{1010}$ are connected to $K_{1009}$. One can proved that this configuration will only get to cover $K_{a} K_{2019 - a}$ by inverting any vertex. Now, we will prove that this is the upper bound. Claim 01. We can prove that every vertex has degree at most 1009. Proof. Suppose there exists a vertex with degree more than 1009, by inverting that vertex, we could assure that the number of edges strictly decreases. So, at one point, we can assure that every vertex has degree at most 1009. Claim 02. There mustn't be more than 1010 points of degree 1009. Proof. Suppose not, then there exists a pair of vertex $(A,B)$ connecting to each other, therefore by inverting on $A$, degree of $B$ will rise on to $1010$. Inverting that vertex will strictly decrease the number of edge again. Therefore, \[ 2E = \sum_{i \in G} d(i) \le 1010 \times 1009 + 1009 \times 1008 = 2 \cdot 1009^2 \]Giving us, $E \le 1009^2$.

In Claim 02, it should read: there exists a pair of such vertices (degree 1009) A, B which are NOT connected. Then inverting in A leaves the total number of edges unchanged but increases to the degree of B. Then inverting in B decreases the total number of edges.