Assuming that you mean that for each **round** you generate a new set of **groups**, and everybody in that group plays everybody else in the group:

In round 1, we might as well put teams 1-4 in a group, and ditto for 5-8, 9-12, and 13-16 (otherwise, just renumber the teams).

In round 2, each group has to consist of one team from each round 1 group, so (again, renumbering as necessary), we might as well have 1,5,9,13 | 2,6,10,14 | 3,7,11,15 | 4,8,12,16.

In round 3, each group has to consist of one team from each round 1 group, and also one team from each round 2 group.

Up to relabelling, we might as well put 1 and 6 together. Up to relabelling, we might as well put 11 in there as well. 16 is then the only remaining team that hasn't been with 1, 6, or 11, so we've got to put them in as well.

Now, 2 must be in a group with one of 5 and 7, one of 9 and 12, and one of 13 and 15 (round 1), and also exactly one of 5,9,13, one of 7 and 15, and 12. Thus, 2 must be with 12, so can't be with 9. Thus, we can either have 5,15 or 7,13 in there: up to relabelling, it doesn't matter which, so we'll go 5,15.

Now, 3 must be with one of 7 and 8, one of 9 and 10, and one of 13 and 14, and not 7, and also one of 9 and 13, one of 10 and 14, and 8, so 3 must be with 8, so can't be with 7, so our group is either 3,8,9,14 or 3,8,10,13, and, up to relabelling, it doesn't matter which, so we'll go with 3,8,9,14.

Now, the only remaining teams are 4, 7, 10, and 13, and that is, indeed, a valid group, so we can fill the third round.

But now, in the fourth round, 1 must be in a group with one of 5,6,7,8, one of 9,10,11,12, and one of 13,14,15,16 (round 1), and also one of 2,6,10,14, one of 3,7,11,15, and one of 4,8,12,16 (round 2), and one of 2,5,12,15, one of 3,8,9,14, and one of 4,7,10,13. But 1 can't be with 2,3,4,5,6,9,10,13, or 16, so taking those out of the lists, 1 must be in a group with one of 7,8, one of 11,12, and one of 14,15, (round 1), 14, one of 7,11,15, and one of 8,12 (round 2), one of 12,15, one of 8,14, and 7 (round 3), so 1 must be with 7 and 14, so can't be with 8, 11 or 15, so our final team must be 12 (by round 2), so 1,7,12,14 is one of our groups

And 2 must be with one of 5,6,8, one of 9,10,11, one of 13,15,16 (R1), one of 5,9,13, one of 3,11,15, one of 4,8,16 (R2), one of 6,11,16, one of 3,8,9, and one of 4,10,13 (R3). But 2 can't be with 3,4,5,6,10,12,14,15, so taking those out, we have that 2 must be with 8, one of 9,11, one of 13,16 (R1), one of 9,13, team 11 (R2), one of 11,16, one of 8,9, team 13 (R3), so 2,8,11,13 is another one of our groups.

Now, 3 must be with one of 5,6, one of 9,10, one of 15,16 (R1), one of 5,9, one of 6,10, one of 4,16 (R2), one of 6,16, one of 5,15, and one of 4,10 (R3). But 3 has already been with 1,2,4,7,8,9,11,14,15, so taking those out, 3 must be with 6, 10, 16 (R1), but also with 5 (R2), a contradiction, so this fourth round is not possible.

## Fedzbar

If I understood correctly, there are 4 teams in each group. Clearly this means that each team in this group has 3 teams to play against, so there can only be 3 rounds.