My answer is $2k-1$. Suppose there are at least $2k$ rounds then in the last round we match two friends who were never paired before, this means that in each previous round at least one of them was paired with someone else (who is not the friend in the last round) then before this there were at most $(k-1) + (k-1)$ rounds so there are at most $2k-1$ rounds (adding the last round). For the example we see it as a graph: Take two vertices $ A $, $ B $ where $ A $ is joined to $ v_1 \dots v_k $ and $ B $ is joined to $ u_1 \dots u_k $ where the $2k + 2$ vertices are different two by two, and also each $ u_i $ is linked with $ v_j $ where $ j, i $ are different. Then take the first $ k-1 $ rounds for the edges between $ u_i, v_{i + t} $ with $t$ is the number of the round ($t = 1, \dots, k-1$), now for the last $ k $ rounds only take in each round the edges $ Av_i $ and $ Bu_i $. This is correct?