We need to find all integers $n \ge 2$ for which there exists an integer $m$ and a polynomial $P(x)$ with integer coefficients satisfying the following conditions:
1. $m > 1$ and $\gcd(m, n) = 1$;
2. The numbers $P(0), P^2(0), \ldots, P^{m-1}(0)$ are not divisible by $n$;
3. $P^m(0)$ is divisible by $n$.

### Part A: Necessity

We start by proving a critical lemma about cycles modulo $p^t$.

Lemma: Consider the mapping from $\mathbb{Z}/p^t\mathbb{Z}$ to itself defined by $x \mapsto P(x)$. Then the length of each cycle in this mapping must be $p$-smooth (i.e., each prime factor does not exceed $p$).

Proof: We use induction on $t$. The base case $t = 1$ is obvious as the length of each cycle is $1, 2, \ldots, p$. Assume this is true for $t-1$. Consider a cycle
$$a \to P(a) \to P(P(a)) \to \cdots \to P^{(n)}(a) = a$$
and let $m$ be the smallest integer such that $p^{t-1} \mid P^{(m)}(a) - a$ (i.e., the length of the cycle viewed modulo $p^{t-1}$). Clearly, $m \mid n$ and $m$ is $p$-smooth. Let $T(x) = P^{(m)}(x)$. Then $T(a), T(T(a)), \ldots$ must be congruent to $a \pmod{p^{t-1}}$. There are $p$ such residues that are congruent to $a \pmod{p^{t-1}}$, so $m/n \in \{1, 2, \ldots, p\}$, which implies the claim. $\blacksquare$

Now, order all primes $p_1 < p_2 < \cdots$ in increasing order and suppose that $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ works where $a_1, a_2, \ldots, a_k > 0$. Since $m > 1$, there exists an index $i$ such that the length $\ell$ of the cycle of $P$ modulo $p_i^{a_i}$ that contains $0$ is $> 1$. However, by the lemma, $\ell$ must be $p_i$-smooth, which is forced to be not coprime to $n$, a contradiction.

### Part B: Sufficiency

Let $p$ be the largest prime divisor of $n$ and $t = \nu_p(n)$. Due to the condition on $n$, there exists a prime $q < p$ that does not divide $n$. Take $m = q$. First, we construct this polynomial modulo $p^t$.

Claim: There exists a polynomial $Q \in \mathbb{Z}[x]$ such that the length of the cycle containing $0$ modulo $p^t$ is exactly $q$.

Proof: Take any integers $x_1, x_2, \ldots, x_{q-1}$ of your choice such that they are distinct modulo $p$. Set $x_0 = x_q = 0$. Take the Lagrange Interpolation polynomial
$$Q(x) = \sum_{i=0}^{q-1} x_{i+1} \left( \prod_{\substack{j \ne i \\ j \in \{0, 1, \ldots, q-1\}}} \frac{x - x_j}{x_i - x_j} \right) \pmod{p^t}$$
so that $Q(x_i) = x_{i+1}$ for all $i = 0, 1, 2, \ldots, q-1$. Fractions are acceptable since the denominators are not divisible by $p$, so we use inverses modulo $p^t$. $\blacksquare$

Take $Q$ from above and let $Q(x) = \sum_{i=0}^{d} a_i x^i$. We construct $b_i$ such that
$$\begin{align*} b_i &\equiv 0 \pmod{\frac{n}{p^t}}, \\ b_i &\equiv a_i \pmod{p^t}, \end{align*}$$
and let $P(x) = \sum_{i=0}^d b_i x^i$. It's clear that modulo $\frac{n}{p^t}$, $P$ maps $0$ to itself, while modulo $p^t$, it takes $q$ applications of $P$ to get back to $0$. Hence $m = q$ works and we are done.

Thus, the answer is: $n$ works if and only if the set of prime divisors of $n$ is not the set of the first $k$ primes for some $k$.

The answer is: $\boxed{n \text{ works if and only if the set of prime divisors of } n \text{ is not the set of the first } k \text{ primes for some } k}$.