Four. (50 points) Given that there are $n(n \geqslant 4)$ football teams participating in a round-robin tournament, each pair of teams plays one match. The winning team gets 3 points, the losing team gets 0 points, and in case of a draw, both teams get 1 point. After all the matches, it is found that the total scores of the teams form an arithmetic sequence with a common difference of 1. Find the maximum score of the last place team.

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All teams play a total of $\mathrm{C}_{n}^{2}=\frac{n(n-1)}{2}$ matches.
Let the score of the last place be $k$. Then the total score is
$$
S=\sum_{i=0}^{n-1}(k+i)=\frac{n(2 k+n-1)}{2} .
$$

If a team has no draws, then each match score is either 0 or 3, so the total score is a multiple of 3.

Given that the total scores of all teams form an arithmetic sequence, there must be draws, i.e., it is not possible for both teams to score a total of 3 points in every match. Hence,
$$
\begin{array}{l}
\frac{n(2 k+n-1)}{2}<3 \times \frac{n(n-1)}{2} \\
\Rightarrow k<n-1 \Rightarrow k \leqslant n-2 .
\end{array}
$$

We will prove by induction on $n$ that $k=n-2$ can be achieved.
Let $A(k)$ denote the team with a score of $k$.
When $n=4$, let team $A(5)$ defeat $A(4)$, team $A(4)$ defeat $A(2)$, and all other matches end in draws. Then $A(5)$ has one win and two draws, totaling 5 points; $A(4)$ has one win, one draw, and one loss, totaling 4 points; $A(3)$ has three draws, totaling 3 points; and $A(2)$ has two draws and one loss, totaling 2 points.

Assume that for $n$ teams, scores $A(n-2), A(n-3), \cdots, A(2 n-3)$ can be achieved. Add a new team $C$ to make the scores $n-1, n-2, \cdots, 2 n-1$. Let $n=3 q+r$ where $r \in\{0, \pm 1\}$.
Let team $C$ win $n-1-q$ matches, draw one match, and lose $q$ matches. Then its score is
$$
1+3(n-1-q)=3 n-3 q-2=2 n-2+r \text {. }
$$

The adjustment strategy is for team $C$ to lose to $A(k), A(k+3), \cdots, A(k+3 q-3)$ to move them up, and then use one draw to fine-tune.
(1) If $r=1$, team $C$ scores $2 n-1$. Let team $C$ lose to $A(n-1), A(n+2), \cdots, A(2 n-5)$, making $n-1$ become $2 n-2$, and then draw with $A(n-2)$ to make it rise to $n-1$.
(2) If $r=0$, team $C$ scores $2 n-2$. Let team $C$ lose to $A(n-1), A(n+2), \cdots, A(2 n-4)$, making $n-1$ become $2 n-1$, and then draw with $A(n-2)$ to make it rise to $n-1$.
(3) If $r=-1$, team $C$ scores $2 n-3$. Let team $C$ lose to $A(n-2), A(n+1), \cdots, A(2 n-4)$, making $n-2$ become $2 n-1$, and then draw with $A(2 n-3)$ to make it rise to $2 n-2$.
Therefore, the maximum score for the last place is $n-2$.