2N-4 is looking good as a minimal solution. (I can't come up with an 11 call solution for N=8, either.)

If 2N-4 is minimal, then a solution, or rather two solutions, to your "bookkeeping" twist are strongly suggested by the forms of the inductive arguments you noted. With the tiniest bit of tidying up you can turn either into an algorithm. As long as each person knows his part in the algorithm, he doesn't need to know who has called whom or what anyone else knows.