Find all polynomials $W$ with integer coefficients satisfying the following condition: For every natural number $n, 2^n - 1$ is divisible by $W(n).$
1. We need to find all polynomials \( W \) with integer coefficients such that for every natural number \( n \), \( W(n) \) divides \( 2^n - 1 \). This can be written as:
   \[
   W(n) \mid 2^n - 1 \quad \forall n \in \mathbb{N}
   \]

2. First, consider the degree of the polynomial \( W(n) \). Let \( \deg(W) = d \). If \( d > 1 \), then \( W(n) \) grows faster than \( 2^n - 1 \) as \( n \) increases, which would make it impossible for \( W(n) \) to divide \( 2^n - 1 \) for all \( n \). Therefore, \( d \leq 1 \).

3. If \( d = 0 \), then \( W(n) \) is a constant polynomial, say \( W(n) = a \). For \( W(n) \) to divide \( 2^n - 1 \) for all \( n \), \( a \) must be a divisor of \( 2^n - 1 \) for all \( n \). The only integers that divide \( 2^n - 1 \) for all \( n \) are \( \pm 1 \). Thus, \( W(n) = 1 \) or \( W(n) = -1 \).

4. If \( d = 1 \), then \( W(n) \) is a linear polynomial, say \( W(n) = an + b \). We need \( an + b \mid 2^n - 1 \) for all \( n \). Consider the case when \( n = 1 \):
   \[
   a \cdot 1 + b \mid 2^1 - 1 \implies a + b \mid 1
   \]
   The divisors of 1 are \( \pm 1 \). Therefore, \( a + b = 1 \) or \( a + b = -1 \).

5. If \( a + b = 1 \), then \( b = 1 - a \). Thus, \( W(n) = an + (1 - a) \). Simplifying, we get:
   \[
   W(n) = an + 1 - a
   \]
   If \( a = 2 \), then \( W(n) = 2n - 1 \).

6. If \( a + b = -1 \), then \( b = -1 - a \). Thus, \( W(n) = an + (-1 - a) \). Simplifying, we get:
   \[
   W(n) = an - 1 - a
   \]
   If \( a = -2 \), then \( W(n) = -2n - 1 \).

7. Therefore, the polynomials \( W(n) \) that satisfy the given condition are:
   \[
   W(n) = 1, \quad W(n) = -1, \quad W(n) = 2n - 1, \quad W(n) = -2n + 1
   \]

The final answer is \(\boxed{W \in \{1, -1, 2n - 1, -2n + 1\}}\).