IMO Shortlist

For $n = 1, 2, \ldots$ define
$$d_{n} = \left(z_{0} + z_{1} + \cdots + z_{n}\right) - n z_{n}$$
The sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n} > 0$.
Notice that
$$n z_{n+1} - \left(z_{0} + z_{1} + \cdots + z_{n}\right) = (n+1) z_{n+1} - \left(z_{0} + z_{1} + \cdots + z_{n} + z_{n+1}\right) = -d_{n+1},$$
so the second inequality in (1) is equivalent to $d_{n+1} \leqslant 0$. Therefore, we have to prove that there is a unique index $n \geqslant 1$ that satisfies $d_{n} > 0 \geqslant d_{n+1}$.
By its definition the sequence $d_{1}, d_{2}, \ldots$ consists of integers and we have
$$d_{1} = \left(z_{0} + z_{1}\right) - 1 \cdot z_{1} = z_{0} > 0 .$$
From
$d_{n+1} - d_{n} = \left(\left(z_{0} + \cdots + z_{n} + z_{n+1}\right) - (n+1) z_{n+1}\right) - \left(\left(z_{0} + \cdots + z_{n}\right) - n z_{n}\right) = n\left(z_{n} - z_{n+1}\right) < 0$
we can see that $d_{n+1} < d_{n}$ and thus the sequence strictly decreases.
Hence, we have a decreasing sequence $d_{1} > d_{2} > \ldots$ of integers such that its first element $d_{1}$ is positive. The sequence must drop below 0 at some point, and thus there is a unique index $n$, that is the index of the last positive term, satisfying $d_{n} > 0 \geqslant d_{n+1}$.