Robin Round Tournament 3: Min Number to Pass Through

There are a total of \(\binom{10}{2} = 45\) games. First, assign as many wins as possible to the top 3 teams, say they have \(9, 8,\) and \(7\) wins respectively (this sums to \(24\) wins). That leaves \(45 - 24 = 21\) wins. We want these wins to be distributed equally among the remaining 7 teams for obvious reasons. If those 21 wins are spread evenly, each of the other 7 teams ends up with 3 wins. This creates a 7-way tie for 4th place, each with 3 wins. In principle, any one of those tied teams could advance (via a lottery). Thus, a team could qualify for the top 4 with as few as \(\boxed{3}\) wins.