I think the parent was agreeing with you, but saying that the number of edges on the fully connected (usually called complete) graph on n vertices is 1 + 2 + 3 + … + n = n(n + 1)/2.

Whenever you visit a node you need one edge to walk in and one to walk out, except for the first and last node. So all K_n with even n , you can walk all the edges. For odd n, you lose one edge per node, except for 2 of them