Yes, for all $n$, the edge-chromatic number of $K_{2n}$ is $2n-1$ and the edge chromatic number of $K_{2n+1}$ is $2n+1$.  Moreover, it is easy to construct such edge-colourings in polynomial time.  For example, see this [Wikipedia][1] page for an easy method to construct a "$1$-factorization" of $K_{2n}$.  An optimal edge-colouring for $K_{2n+1}$ can be obtained by applying this method to $K_{2n+2}$, and then deleting the edges incident to the extra vertex.  


  [1]: https://en.wikipedia.org/wiki/Graph_factorization