We prove this with two-step induction to $n$. For $n=2$, it follows from the conditions that $z_1 = z_2 = 1$, and so then $z_1 - z_2 = 0$.

For $n=3$, it follows from the conditions that $(z_1, z_2, z_3)$ equals $(1, 1, 2)$, $(1, 2, 1)$, or $(1, 2, 3)$. In these cases, we choose $(s_1, s_2, s_3)$ to be equal to $(1, 1, -1)$, $(1, -1, 1)$, and $(1, 1, -1)$, respectively.

Now suppose we have proved the statement for $n=k$ and $n=k-1$, where $k \ge 2$. Let $z_1, \dots, z_{k+1}$ be numbers that satisfy the conditions in the problem statement. We distinguish two cases.

First, suppose that $z_k = z_{k+1}$. Applying the induction hypothesis for $n=k-1$ to $z_1, \dots, z_{k-1}$ we find $s_1, \dots, s_{k-1}$ such that $s_1z_1 + \dots + s_{k-1}z_{k-1} = 0$. Now we also have that

$$s_1z_1 + \dots + s_{k-1}z_{k-1} + z_k - z_{k+1} = 0,$$
which proves the statement in this case.

Now suppose that $z_k \neq z_{k+1}$. Then $|z_k - z_{k+1}| \ge 1$. In addition, we also know that $|z_k - z_{k+1}| \le k$ because $1 \le z_k, z_{k+1} \le k+1$. Since $|z_k - z_{k+1}| = z_k + z_{k+1} \pmod 2$, we also have that $z_1 + \dots + z_{k-1} + |z_k - z_{k+1}|$ is even. So we can apply the induction hypothesis for $n=k$ to $z_1, \dots, z_{k-1}, |z_k - z_{k+1}|$. As a result, we find $s_1, \dots, s_k$ such that:
$$s_1z_1 + \dots + s_{k-1}z_{k-1} + s_k|z_k - z_{k+1}| = 0.$$
Since $|z_k - z_{k+1}| = \pm (z_k - z_{k+1})$, this proves the statement. $\square$