Problem 1848

a) The total number of matches is $n(n-1)/2$. Let $w$ be the number of games ended with a victory and $e$ the number of games ended in a draw ($e \ge 1$ due to the last team). Thus, $w+e = n(n-1)/2$. If $r$ is the step of the arithmetical progression, we have that the total number of points in the final classification is
$$\frac{(2 + (n-1)r)n}{2} = 3w + 2e = w + 2(w + e) = w + n(n-1),$$
or
$$2w - 2n = n(n-1)(r-2).$$
The case $r=0$ is obviously impossible (each team should have only 1 point). For $r=1$ we get $2w = 3n - n^2$, a contradiction (for $n=3$ one has 3 games, $w=0$, all games ended in a draw, but the last team has only 1 point; the case $n \ge 3$ implies $w < 0$). If $r \ge 3$, then $2w \ge 2n + n(n-1) = n(n+1)$ in contradiction with the fact that the total number of matches is $n(n-1)/2$.
Thus, the only possible value is $r=2$. In this case $w=n$ and the number of points of each team in decreasing order is the sequence $2n-1, 2n-3, 2n-5, \dots, 1$.
Denote by $w_i$ and by $e_i$ the number of victories and draws of the $i$-th classified team (in decreasing order). Note that $w_i + e_i \le n-1$. Considering the number of points obtained by the first three teams, that is $2n-1, 2n-3, 2n-5$, we get $2n-1 = 3w_1+e_1 = 2w_1+w_1+e_1 \le 2w_1+n-1$, that is $w_1 \ge n/2$. Analogously, $w_2 \ge n/2-1$ and $w_3 \ge n/2-2$.
For $n \ge 7$ we obtain that $w_1 + w_2 + w_3 \ge 3n/2 - 3 > n$, a contradiction with the fact that the total number of victories is $n$. It implies that $n \le 6$ and it is impossible to have $n=12$.

b) It is clear that such a configuration does not exist for $n=3$ ($w=3, e=n(n-1)/2-w=0$).
The case $n=4$ is possible, where the points $7, 5, 3, 1$ in the final classification are realized by the following results of the matches of teams: T1-T2 (a draw), T1-T3 (T1 won), T1-T4 (T1 won), T2-T3 (T2 won), T2-T4 (a draw), T3-T4 (T3 won).

In this case we have to write 7 points of the third as a sum of at most five numbers of 1's, which is impossible.
Therefore, the only possibility is $n=4$, the configuration being described above. ☐