Answer: $4,8,12,16$.

Solution. Note that the girls participated in a total of $4+6+7+10+11=38$ games, with two girls participating in each game, so there were 19 games in total. The key insight in the solution is that a girl cannot skip four games in a row. Therefore, Anya could only play 4 games in the following way: she did not play 3 times, then played, did not play 3 times, then played, and so on. Thus, Anya played in games numbered 4, $8,12,16$, and lost all of them.

Comment. The girls could indeed play as described in the problem: B-G, V-G, D-G, A-G, B-G, V-G, G-D, A-D, B-D, V-D, G-D, A-D, D-B, B-V, $\quad \Gamma-\mathrm{B}, \quad \mathrm{A}-\mathrm{B}, \quad \mathrm{B}-D, \quad$ D-B, D-G.
6-8. Five girls played several table tennis matches at one table. At any given time, two of the girls played, while the other three rested. The girl who lost a match went to rest, and her place at the table was taken by the girl who had rested the most; if there were several such girls, any one of them could take the place. There were no ties in the tennis matches. Anya played 4 matches, Bella - 6 matches, Valya - 7 matches, Galia - 10 matches, and Dasha - 11 matches. Indicate the numbers of all the matches in which Anya lost.