A positive integer $n$ is [i]friendly[/i] if the difference of each pair of neighbouring digits of $n$, written in base $10$, is exactly $1$. [i]For example, 6787 is friendly, but 211 and 901 are not.[/i]

Find all odd natural numbers $m$ for which there exists a friendly integer divisible by $64m$.
1. **Define Friendly Numbers**: A positive integer \( n \) is friendly if the difference between each pair of neighboring digits of \( n \), written in base 10, is exactly 1. For example, 6787 is friendly, but 211 and 901 are not.

2. **Define Happy Numbers**: We need to find all odd natural numbers \( m \) for which there exists a friendly integer divisible by \( 64m \). We call such a number \( m \) *happy*.

3. **Initial Claim**: We claim that all happy integers are those for which \( (m, 5) = 1 \). This means \( m \) should not be divisible by 5.

4. **Divisibility by 64**: Note that \( 64 \mid 343232 \). This can be verified by checking the divisibility of 343232 by 64:
   \[
   343232 \div 64 = 5363 \quad \text{(an integer)}
   \]
   Hence, \( 343232 \) is divisible by 64.

5. **Contradiction for \( 5 \mid m \)**: If \( 5 \mid m \) and \( m \) is happy, then if \( n \) is friendly and a multiple of \( 64m \), \( n \) must end in either 0 or 5.
   - If \( n \) ends in 0, the second to last digit must be 1 (since the difference between neighboring digits is 1). Thus, \( n \) ends in 10, which is not a multiple of 4, leading to a contradiction.
   - If \( n \) ends in 5, then \( n \) is odd, and thus not even divisible by 2, leading to another contradiction.

6. **Case for \( (m, 5) = 1 \)**: Suppose \( (m, 5) = 1 \). Let \( n \) consist of \( t \) consecutive strings of \( 343232 \). Then,
   \[
   n = 343232 \cdot \frac{10^t - 1}{9}
   \]
   Since \( 64 \mid 343232 \), for \( 64m \mid n \) to hold, we need \( 9m \mid 10^t - 1 \).

7. **Using Euler's Totient Function**: Since \( (m, 5) = 1 \) and \( m \) is odd, we have \( (9m, 10) = 1 \). By Euler's theorem, \( 10^{\varphi(9m)} \equiv 1 \pmod{9m} \), where \( \varphi \) is Euler's totient function. Therefore, we can pick \( t \) to be any multiple of \( \varphi(9m) \).

8. **Conclusion**: Since we can always find such a \( t \) for any \( m \) with \( (m, 5) = 1 \), we conclude that all odd natural numbers \( m \) for which there exists a friendly integer divisible by \( 64m \) are those for which \( (m, 5) = 1 \).

The final answer is \( \boxed{ m } \) such that \( (m, 5) = 1 \).