1. Assume the contrary: Suppose there exists a natural number that moves infinitely many times. Let this number be in position $p$.

2. Operation description: After $m$ minutes, the series is divided into blocks of $m$ numbers. The first block remains unchanged. In each subsequent block, the first number moves to the end, and all other numbers shift one place to the left.

3. Position update rule: If the number is in position $p$ at minute $m$:
- If $m \mid p$, then $p$ is replaced by $p + m - 1$.
- Otherwise, $p$ is replaced by $p - 1$.

4. Infinite movement implication: Since the number moves infinitely often, it can never reach $p = 0$. Therefore, it must move to the right infinitely often.

5. Movement analysis: Suppose at minute $N+1$, the number moves forward from position $p = k(N+1)$ to $p = (k+1)(N+1) - 1$.

6. Non-increasing ratio claim: The ratio $k$ is non-increasing. If the next rightward move occurs at minute $N + \ell + 1$, then:
$$N + \ell + 1 \mid (k+1)(N+1) - \ell - 1$$
This ratio is at most $k$ and exactly $k$ only when $\ell \cdot (k+1) = N$.

7. **Stabilization of $k$**: Since $k$ cannot become zero, it must stabilize at some value $K$.

8. **Largest integer $z$**: Let $z$ be the largest nonnegative integer such that $(K+1)^z \mid N$.

9. **Sequence of $\ell$**: The first value of $\ell$ is $\frac{N}{k+1}$. The next value is:
$$\frac{N + \ell}{k+1} = \frac{N}{k+1} + \frac{N}{(k+1)^2}$$
and so on.

10. Integer condition: These values cannot be integers after $z$ steps, causing $k$ to decrease.

11. Conclusion: Eventually, $k$ becomes smaller than one, at which point the number stops moving.

Therefore, each natural number will only move a finite number of times.

$\blacksquare$