Find all polynomials $p(x)$ with integer coefficients such that for each positive integer $n$, the number $2^n - 1$ is divisible by $p(n)$.
1. **Assume \( p(x) \) is a polynomial with integer coefficients such that \( p(n) \) divides \( 2^n - 1 \) for all positive integers \( n \).**

2. **Consider a prime \( q \) that divides \( p(n) \) for some integer \( n \). Then \( q \) must also divide \( 2^n - 1 \).**

3. **Since \( q \) divides \( p(n) \), it must also divide \( p(n+q) \) because \( p(x) \) has integer coefficients and thus \( p(n+q) \equiv p(n) \pmod{q} \).**

4. **Therefore, \( q \) must divide \( 2^{n+q} - 1 \). Notice that:**
   \[
   2^{n+q} - 1 = 2^n \cdot 2^q - 1 = 2^n \cdot 2^q - 2^n + 2^n - 1 = 2^n (2^q - 1) + (2^n - 1)
   \]
   Since \( q \) divides \( 2^n - 1 \), it must also divide \( 2^{n+q} - 1 \).

5. **Thus, \( q \) must divide \( 2^q - 1 \). By Fermat's Little Theorem, for any prime \( q \), \( 2^q \equiv 2 \pmod{q} \), which implies \( 2^q - 1 \equiv 1 \pmod{q} \). Therefore, \( q \) must divide \( 1 \), which is a contradiction unless \( q = 1 \).**

6. **Since \( q \) is a prime, the only possibility is that \( q = 1 \), which is not a valid prime. Therefore, \( p(n) \) must be a constant polynomial.**

7. **If \( p(n) \neq \text{const} \), we can use Schur's theorem, which states that a non-constant polynomial with integer coefficients takes on infinitely many prime values. This would contradict the fact that \( q \) must divide \( 2^n - 1 \) for all \( n \).**

8. **Thus, the only polynomial \( p(x) \) that satisfies the given condition is a constant polynomial.**

9. **Let \( p(x) = c \), where \( c \) is an integer. Then \( c \) must divide \( 2^n - 1 \) for all positive integers \( n \).**

10. **The only integer \( c \) that divides \( 2^n - 1 \) for all \( n \) is \( c = 1 \).**

The final answer is \( \boxed{ p(x) = 1 } \).