$n(n \geqslant 4)$ football teams participate in a round-robin tournament, where 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 the tournament, it is found that the total scores of all teams form an arithmetic sequence with a common difference of 1. Find the maximum score of the last place team.
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 $S=\sum_{i=0}^{n-1}(k+i)=$ $\frac{n(2 k+n-1)}{2}$.

If a team has not played any draws, then its score per match is either 0 or 3, so the total score is a multiple of 3. From the given score sequence, there must be draws, meaning not every match results in a total of 3 points. Therefore,
$$
\frac{n(2 k+n-1)}{2}<3 \cdot \frac{n(n-1)}{2},
$$

Solving this, we get $k<n-1$, i.e., $k \leqslant n-2$.
We will prove by induction on $n$ that $k=n-2$ can be achieved.
Let $A(k)$ denote the team with $k$ points. When $n=4$, let $A(5)$ defeat $A(4)$, $A(4)$ defeat $A(2)$, and all other matches end in draws. Then $A(5)$ has 1 win and 2 draws, totaling 5 points; $A(4)$ has 1 win, 1 loss, and 1 draw, totaling 4 points; $A(3)$ has 3 draws, totaling 3 points; $A(2)$ has 1 loss and 2 draws, totaling 2 points.

Assume for $n$ teams, the scores can be $n-2, \cdots, 2 n-3$, with the corresponding teams being $A(n-2), \cdots, A(2 n-3)$. Add a new team $C$, aiming to achieve scores of $n-1, \cdots, 2 n-1$. Let $n=3 q+r, r \in\{0, \pm 1\}$. Let $C$ have 1 draw, $q$ losses, and $n-1-q$ wins, then its score is
$$
1+3(n-1-q)=3 n-3 q-2=2 n-2+r .
$$

The adjustment strategy is for $C$ 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 $r=1, C$'s score is $2 n-1$, let $C$ lose to $A(n-1), A(n+2), \cdots, A(2 n-5)$, changing $n-1$ to $2 n-2$, and then draw with $A(n-2)$ to move it up to $n-1$.
(2) If $r=0, C$'s score is $2 n-2$, let $C$ lose to $A(n-1), A(n+2), \cdots, A(2 n-4)$, changing $n-1$ to $2 n-1$, and then draw with $A(n-2)$ to move it up to $n-1$.
(3) If $r=-1, C$'s score is $2 n-3$, let $C$ lose to $A(n-2), A(n+1), \cdots, A(2 n-4)$, changing $n-2$ to $2 n-1$, and then draw with $A(2 n-3)$ to move it up to $2 n-2$.
Therefore, the maximum score of the last place is $n-2$.