I had posted the problem because I couldn't see how to solve it for some time but for some strange reason I found a simple answer not long after I put it on MO, so it is rather to be seen as a comment:
The basic idea for constructing an extremal example is to take a densest triangle-free graph and set its edgeweights to $1$ and augment it to a complete graph by adding edges of weight $3$.
Densest triangle-free graphs are $K_{n,n}$ with $n^2$ edges, implying that the number of non-metric edges that augment it to $k_{2n}$ is $\ n\cdot(2n-1)-n^2\ =\ n^2-1$ if the number of vertices is $2n$, which amounts to $\approx 25\%$