Team "A" is the winner when 2 points are calculated for a victory, and takes the last place when 3 points are calculated for a victory, while for the team "B" it is all the way around. Let the team "A" win $x$ matches and have $y$ draws. Team "B" will have $a$ wins and $b$ draws, respectively. Since there could not be only two teams, the rest of the teams gained the numbers of points that between the results of teams "A" and "B" in both cases of calculation. So we have the following inequalities:
$$2x + y \ge 2a + b + 2 \text{ and } 3x + y + 2 \le 3a + b \Rightarrow 2(a - x) + 2 \le y - b \le 3(a - x) - 2.$$
Consider the possible values of $(y-b)$ for different values of $(a-x)$.
For $a-x=0$ we have that $2 \le y-b \le -2$ which is impossible.
For $a-x=1$ we have that $4 \le y-b \le 1$ which is impossible.
For $a-x=2$ we have that $6 \le y-b \le 4$ which is impossible.
For $a-x=3$ we have that $8 \le y-b \le 7$ which is impossible.
For $a-x=4$ we have that $10 \le y-b \le 10$.

Thus, the minimum possible value for $y-b=10$. So either $y-b \ge 10$ or $y \ge 10+b$. Thus the smallest value for $y=10$. Since it is clear that $x \ge 1$, then $x+y \ge 11$, i.e. the should be at least 12 teams.

What is left to show that for 12 teams there exists such distribution of points. In the table (fig. 41) victory is denoted by 1, defeat by 0 and empty cells correspond to draws between the teams.
Then in the championship when calculating by the rule "2 points for a win" team "A" (number 2 in the table) gains 12 points, "B" (number 1 in the table) gains 10 points, and the rest of the teams gain 11 points each.
In the championship when calculating by the rule "3 points for a win" team "A" gains 13 points, the team "B" gains 15 points, and the rest of the teams gain 14 points each.

Fig. 41