4.084 In a volleyball competition, $n$ teams participated. Each team played with all the others once. For each game, the winning team was awarded one point, and no points were awarded for a loss, as there are no draws in volleyball. At the end of the competition, it turned out that the points scored by the teams formed an arithmetic progression. How many points did the team that finished in last place score?
Solution. Let the points scored by the teams form a non-decreasing arithmetic progression $a_{1}, a_{2}, \ldots, a_{n}$ with a common difference $d$. A total of $\frac{n(n-1)}{2}$ games were played. Therefore, the total number of points scored by all teams is $\frac{n(n-1)}{2}$, from which we have $a_{1}+a_{2}+\ldots+a_{n}=\frac{n(n-1)}{2} \Rightarrow \frac{n}{2}\left(2 a_{1}+(n-1) d\right)=\frac{n(n-1)}{2} \Rightarrow 2 a_{1}=(n-1)(1-d)$.

Since $d$ and $a_{1}$ are integers, we have $\left[\begin{array}{l}d=1, \\ d=0 .\end{array}\right.$ However, according to the problem, there is a team that finished in the last place, so $d=1 \Rightarrow a_{1}=0$.

Answer: 0.