Consider the problem of finding all positive integers $n$ such that there exists a polynomial $P(x) \in \mathbb{Z}[x]$ meeting the specified condition: for every positive integer $m \geq 1$, the sequence $P^m(1), P^m(2), \ldots, P^m(n)$ produces exactly $\left\lceil \frac{n}{2^m} \right\rceil$ distinct remainders when divided by $n$. Here, $P^m$ denotes $P$ iterated $m$ times.

### Step 1: Analyze the Condition

For a given $n$, the problem requires that the application of the polynomial $P$, repeated $m$ times, transforms $1, 2, \ldots, n$ into numbers producing specified distinct residues modulo $n$.

### Step 2: Consider the Case where $n$ is a Prime

1. If $n$ is a prime, then the polynomial $P$ might simplify structuring on $\mathbb{Z}/n\mathbb{Z}$, potentially allowing $P(x) \equiv x^k \mod n$ to have the necessary property of splitting the image set into exactly $\left\lceil \frac{n}{2^m} \right\rceil$ different values for any iteration $m$.
2. Since $n$ is prime, every non-zero residue in $\mathbb{Z}/n\mathbb{Z}$ can appear up to $n - 1$ times. Such behavior aligns well with producing the required distinct remainders when compiled and reduced by powers of 2, as shown by ceiling divisions.

### Step 3: Consider the Case where $n = 2^k$

1. If $n = 2^k$, the binary division by powers of 2 simplifies to subsequent fixed factors. It allows $P(x) \equiv x+c$ (a constant polynomial) to iterate in a manner that naturally breaks into $\left\lceil \frac{2^k}{2^m} \right\rceil$, simplifying into manageable binary expression splits.
2. Each iteration $m$ reduces the effective set size by half, aligning adequately with the required number of distinct residues.

### Conclusion

Analyzing both scenarios, it becomes evident that only when $n$ is either a prime number or a power of 2 can the polynomial $P(x)$ be constructed to satisfy the designated residue conditions for all $m \geq 1$.

Thus, the set of all positive integers $n$ fulfilling the condition are:
$$\boxed{\text{prime } n \text{ and } n = 2^k}$$