There's a counterexample with 8 teams: A, B, C, X, Y, Z, M, S
- S wins against M
- M wins against A, B, and C
- A wins against Y and Z
- B wins against Z and X
- C wins against X and Y
- X wins against M and A
- Y wins against M and B
- Z wins against M and C
- Everything else ties.
Teams ABCXYZ each win twice and lose twice for a score of $2\times0+3\times1+2\times3=9$.
Team M wins against ABC and loses to XYZS, for a score of $4\times0+0\times1+3\times3=9$.
Team S wins once and ties 6 times, for a score of $0\times 0+6\times1+1\times3=9$.
How I found this: First, for easier counting I changed the rules by subtracting $n-1$ points from each team such that ties give $0$ points, and wins and losses each $2$ and $-1$. That makes it easier to see which possible combinations of wins and losses add up to the same. Then I'm looking for a directed graph with no 2-loops such that the value of each node is the same.
Since there are as many wins as losses, in a counterexample there must be at least one team that win more times than they lose. But that team cannot possibly have less than $2$ points, so let's see which way we can make a node with value 2: They are: $1W+0L$, $2W+2L$, $3W+4L$ and so forth -- so if we have one of $1W+0L$ and $3W+4L$ the number of wins/losses add up right.
From there it was just a matter of puzzling out where to add $2W+2L$ nodes to the graph such that we don't need more than one match to take place between the same two teams.
Your approach to the first question is entirely reasonable. Since $5\cdot6=30>28$, it’s clear that $6$ wins guarantee advancement. To show that $5$ do not, we need only show that it’s possible for $5$ teams to win $5$ matches each; the arrangement shown by user73985 works fine and is natural enough that it’s the first one that I found as well. (You can complete it by distributing the remaining $3$ wins within the group consisting of teams $6,7$, and $8$; for example, you could have team $6$ beat teams $7$ and $8$, and team $7$ beat team $8$.)
For the second question, suppose that a team has $3$ wins; is it possible that it could advance to the second round? Suppose that teams $1,2$, and $3$ beat each of the higher-numbered teams: team $1$ beats everybody else, team $2$ beats everyone else except team $1$, and team $3$ beats everyone else except teams $1$ and $2$. That accounts for $18$ wins, leaving $10$ to be determined. Team $4$ beats teams $5,6$, and $7$, for $3$ wins, leaving $7$ wins to still to be distributed amongst teams $5,6,7$, and $8$. We want team $4$, with its measly $3$ wins, to make it to the second round, so we’ll try to split the remaining $7$ wins $2,2,2$, and $1$ amongst teams $5,6,7$, and $8$. Team $4$ did not beat team $8$, so team $8$ beat team $4$ and already has one win; we can give it another against team $5$. Now it must lose to teams $6$ and $7$, so they now have one win apiece. We can finish up by letting team $5$ beat team $6$, team $6$ beat team $7$, and team $7$ beat team $5$. To sum up:
- Team $1$ beats teams $2,3,4,5,6,7,8$.
- Team $2$ beats teams $3,4,5,6,7,8$.
- Team $3$ beats teams $4,5,6,7,8$.
- Team $4$ beats teams $5,6,7$.
- Team $5$ beats team $6$.
- Team $6$ beats teams $7,8$.
- Team $7$ beats teams $5,8$.
- Team $8$ beats teams $4,5$.
Thus, it’s possible to make it into the second round with just $3$ wins.
With only $2$ wins, however, it’s impossible to guarantee getting to the second round. The top three finishers cannot have more than $18$ wins altogether (either $7+6+5$ or $6+6+6$). That leaves $10$ wins amongst the remaining $5$ teams, so either each of the bottom $5$ teams has $2$ wins, or one of them has at least $3$ wins. In neither case is a team with just $2$ wins assured of getting into the second round.
There is some ambiguity in the situation in which each of the bottom $5$ teams wins $2$ matches. (This can happen, e.g., if teams $1,2$ and $3$ all beat each of teams $4,5,6,7$, and $8$, team $4$ beats teams $5$ and $6$, team $5$ beats teams $6$ and $7$, team $6$ beats teams $7$ and $8$, team $7$ beats teams $8$ and $4$, and team $8$ beats teams $4$ and $5$.) In this case the rules as given in the question don’t specify what happens. Some of the possibilities are: that only the top three teams go to the second round; that there is a playoff for the fourth position; that the fourth position is chosen randomly; or that the fourth position is decided by some tie-breaker like goal differential. As long as four teams always move on to the second round, it’s still possible for a team with just $2$ wins in the first round to move on, but if that does happen, other teams with $2$ wins fail to move on.
Best Answer
It's not possible to have more than one team draw all their matches.
It's not possible for more than one team to have six draws and one win (two such teams must have drawn their match, since neither lost). Similarly, it's not possible for more than one team to have six draws and one loss, or for more than one team to have five draws and two wins, or five draws and two losses.
It's possible for the remaining three teams to have five draws, one win and one loss, but this can only happen if none of the matches between these three teams are drawn.
This means that at least six non-draws are required. Actually, all the above can happen simultaneously and you should be able to find a way to achieve this.
[edit to include an example, since another answer claims this is impossible]
Say A draws all their matches (7 points), B beats E but draws all other matches (9 points), C beats D and E but draws the rest (11 points), D loses to C but draws the rest (6 points), E loses to B and C but draws the rest (5 points), F beats G but loses to H (8 points), G beats H but loses to F (8 points), H beats F but loses to G (8 points).
Every pair of teams which drew their match finish on a different number of points. F and G finish on the same number, but didn't draw, and neither did G and H, or F and H.