1983 IMO 第 6 题

(CAN 2) Suppose that $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ are positive integers for which $x_{1}+x_{2}+\cdots+x_{n}=2(n+1)$. Show that there exists an integer $r$ with $0 \leq r \leq n-1$ for which the following $n-1$ inequalities hold: $$\begin{aligned} x_{r+1}+\cdots+x_{r+i} & \leq 2 i+1 & & \forall i, 1 \leq i \leq n-r ; \\ x_{r+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} & \leq 2(n-r+i)+1 & & \forall i, 1 \leq i \leq r-1 . \end{aligned}$$ Prove that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r$.

(CAN 2) 假设 $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ 是正整数，其中 $x_{1}+x_{2}+\cdots+x_{n}=2(n+1)$。证明存在一个整数 $r$，其 $0 \leq r \leq n-1$ 满足下列 $n-1$ 不等式： $$\begin{aligned} x_{r+1}+\cdots+x_{r+i} & \leq 2 i+1 & & \forall i, 1 \leq i \leq n-r ; \\ x_{r+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} & \leq 2(n-r+i)+1 & & \forall i, 1 \leq i \leq r-1 。 \end{aligned}$$ 证明如果所有不等式都是严格的，则 $r$ 是唯一的，否则恰好有两个这样的 $r$。