Special team in a soccer tournament

Question

$N$ teams play in a single round-robin soccer tournament. A game has 3 possible outcomes: team 1 wins, team 2 wins or a draw.

Is it possible that one team achieves more wins than any other team and more draws than any other team? If so, what is the smallest $N$ for which this is possible?

Answer 1

N=6

Team X must win more, and draw more, thus all other teams must lose at least twice (2N-2 losses)
Team X must draw, thus wins at most N-2 times. If no other team wins more than once, they as a group win at most N-1 times.
-> One of the other teams must win at least twice (to get the 2N-2 wins needed for the required losses)
-> Team X must win at least 3 times

Team X must draw more, thus draw against another team, thus (some) other teams draw
-> Team X must draw at least 2 times

This is possible for N=6 (the minimum required for the 5 results), e.g.:
X 2 2 2 1 1
0 X 2 2 0 0
0 0 X 0 2 2
0 0 2 x 2 0
1 2 0 0 x 2
1 2 0 2 0 x

Answer 2

N = 6.
Clearly, the dominating team (call it A) must draw at least twice. Since A needs to get some wins, not all of its matches can be draws, which means all teams have undrawn matches. These will be won by players, not all of which can be A, so some other team will have at least one win, and A must have at least two. In order for A to play four matches, N is at least five.
Proof of impossibility for five: Let A beat B and C and draw against D and E, and let B and C draw against each other to minimine the total number of wins. D and E cannot draw, w.l.o.g D wins. Now D must lose to both B and C, which in turn must both lose to E, giving them two wins, a contradiction.
Example scenario for N = 6: A draws against B,C and beats D,E,F. B beats C,D and loses to E,F. C loses to D and beats E,F, and D,E,F beat each other in a cycle.
A has three wins and two draws, B,C have two wins and a draw each, D,E,F have two wins each.

Answer 3

With N teams there are

a total of $N(N-1)/2$ total games to be played. Any particular team will play $N-1$ games. Assume a team is awarded 2 points for a win, 1 point for a draw, and 0 points for a loss. Thus each game awards a total of 2 points, and there are a total of $N(N-1)$ points to be awarded.

If I draw twice and win the rest of my games, I will get $2(N-3) + 2$ points. The remainder of the points need to be distributed amongst the remaining $N-1$ teams.

If there are fewer than 7 teams, there is no way to allocate the remaining points in such a way that all of the teams would have fewer than my number of wins and my number of draws.

With 7 teams, if I draw twice and win the other four games, I'll have a total of 10 points. There are 32 points left to distribute among the remaining 6 teams, meaning an average of 5.33 points per team. This can be done with 2 teams with 3 wins and 0 draws each, and 4 teams with 2 wins and 1 draw each.

Edited to improve solution:

It actually is possible to allocate the points when N = 6, I made an arithmetical error previously which led me to believe it wasn't. For N = 6, you can draw twice and win three times for a total of 8 points. The remaining 22 points can be distributed between the other 5 teams by having two teams with two wins and a draw each, and three teams with two wins.

For N = 5, if you win 2 and draw 2 you have 6 points. There would be 14 points remaining to distribute among the other four teams, meaning they'd have 3.5 points on average. At least one of the teams would have 4+ points, which is only possible if they have at least two wins or at least two draws.