Problem 1605

[Proof] If the $n$ initial numbers are not all equal, then after $n$ steps, the maximum number in this set must decrease, and the minimum number must increase.

Since the maximum and minimum numbers obtained at each step are integers, after a finite number of steps, a set of equal numbers $(a, a, \cdots, a)$ can certainly be obtained.

Assume that from a set of numbers $z_{1}, z_{2}, \cdots, z_{n}$ that are not all equal, we first obtain a set of identical numbers
$$\frac{z_{1}+z_{2}}{2}=\frac{z_{2}+z_{3}}{2}=\cdots=\frac{z_{n-1}+z_{n}}{2}=\frac{z_{n}+z_{1}}{2}$$

Then the numbers $z_{i}$ are equal at every other position. When $n$ is odd, this is impossible. Therefore, $n$ must be even.
Let $z_{2 i-1}=a, z_{2 i}=b\left(1 \leqslant i \leqslant \frac{n}{2}\right)$. If the array $z_{1}, z_{2}, \cdots, z_{n}$ is directly obtained from the array $y_{1}, y_{2}, \cdots, y_{n}$, then we have
$$\begin{array}{l} \frac{y_{1}+y_{2}}{2}=\frac{y_{3}+y_{4}}{2}=\cdots=\frac{y_{n-1}+y_{n}}{2}=a \\ \frac{y_{2}+y_{3}}{2}=\frac{y_{4}+y_{5}}{2}=\cdots=\frac{y_{n}+y_{1}}{2}=b \end{array}$$

From (1), we get $y_{1}+y_{2}+\cdots+y_{n}=\frac{1}{2}^{n a}$,
From (2), we get $y_{2}+y_{3}+\cdots+y_{n}+y_{1}=\frac{1}{2}^{n b}$,
Thus, $a=b$.
This contradicts the fact that $z_{1}, z_{2}, \cdots, z_{n}$ are not all equal. This proves that the array $a, b, a, b, \cdots, a, b$ is the initial array, hence the conclusion of the problem is established.