An integer $ m > 1$ is given. The inﬁnite sequence $ (x_n)_{n\ge 0}$ is deﬁned by $ x_i\equal{}2^i$ for $ i<m$ and $ x_i\equal{}x_{i\minus{}1}\plus{}x_{i\minus{}2}\plus{}\cdots \plus{}x_{i\minus{}m}$ for $ i\ge m$.
Find the greatest natural number $ k$ such that there exist $ k$ successive terms of this sequence which are divisible by $ m$.
1. **Define the sequence and initial conditions:**
   The sequence \((x_n)_{n \ge 0}\) is defined by:
   \[
   x_i = 2^i \quad \text{for} \quad i < m
   \]
   and
   \[
   x_i = x_{i-1} + x_{i-2} + \cdots + x_{i-m} \quad \text{for} \quad i \ge m.
   \]

2. **Establish the recurrence relation for \(i \ge m\):**
   For \(i \ge m\), we can rewrite the recurrence relation as:
   \[
   x_{i+1} = x_i + x_{i-1} + \cdots + x_{i-m+1}.
   \]
   By subtracting \(x_i\) from both sides, we get:
   \[
   x_{i+1} - x_i = x_i - x_{i-m}.
   \]
   Therefore,
   \[
   x_{i+1} = 2x_i - x_{i-m}.
   \]

3. **Analyze the divisibility by \(m\):**
   Suppose \(x_l, x_{l+1}, \ldots, x_{l+k-1}\) are \(k\) successive terms divisible by \(m\). This means:
   \[
   x_l \equiv 0 \pmod{m}, \quad x_{l+1} \equiv 0 \pmod{m}, \quad \ldots, \quad x_{l+k-1} \equiv 0 \pmod{m}.
   \]

4. **Consider the next term \(x_{l+k}\):**
   Using the recurrence relation:
   \[
   x_{l+k} = 2x_{l+k-1} - x_{l+k-1-m}.
   \]
   Since \(x_{l+k-1} \equiv 0 \pmod{m}\), we have:
   \[
   x_{l+k} \equiv -x_{l+k-1-m} \pmod{m}.
   \]
   For \(x_{l+k}\) to be divisible by \(m\), \(x_{l+k-1-m}\) must also be divisible by \(m\).

5. **Determine the maximum \(k\):**
   To ensure \(x_{l+k-1-m}\) is divisible by \(m\), we need:
   \[
   l+k-1-m \ge l \implies k-1 \ge m \implies k \ge m+1.
   \]
   However, this contradicts the assumption that \(k\) successive terms are divisible by \(m\). Therefore, the maximum \(k\) such that there exist \(k\) successive terms divisible by \(m\) is \(m\).

6. **Conclusion:**
   The greatest natural number \(k\) such that there exist \(k\) successive terms of the sequence which are divisible by \(m\) is \(m\).

The final answer is \(\boxed{m}\)