Determine all natural $ n$ for which there exists natural $ m$ divisible by all natural numbers from 1 to $ n$ but not divisible by any of the numbers $ n \plus{} 1$, $ n \plus{} 2$, $ n \plus{} 3$.
To determine all natural numbers \( n \) for which there exists a natural number \( m \) divisible by all natural numbers from 1 to \( n \) but not divisible by any of the numbers \( n + 1 \), \( n + 2 \), \( n + 3 \), we need to analyze the conditions under which such an \( m \) can exist.

1. **Initial Analysis:**
   - \( m \) must be divisible by the least common multiple (LCM) of all numbers from 1 to \( n \), i.e., \( \text{lcm}(1, 2, \ldots, n) \).
   - \( m \) must not be divisible by \( n + 1 \), \( n + 2 \), and \( n + 3 \).

2. **Condition for \( n + 1 \):**
   - If \( n + 1 \) is not a power of a prime, then \( n + 1 \) can be factored into two coprime factors \( a \) and \( b \) such that \( a, b > 1 \) and \( ab = n + 1 \). Since \( a \) and \( b \) are both less than or equal to \( n \), \( \text{lcm}(1, 2, \ldots, n) \) would be divisible by \( n + 1 \), which contradicts the requirement that \( m \) is not divisible by \( n + 1 \).

3. **Condition for \( n + 2 \) and \( n + 3 \):**
   - Similarly, for \( n + 2 \) and \( n + 3 \) to not divide \( m \), they must also be of the form \( p^k \) for some prime \( p \) and integer \( k \).

4. **Case Analysis:**
   - **Case 1: \( n = 1, 2, 3 \)**
     - For \( n = 1 \), \( m = 1 \) works since 1 is divisible by 1 and not by 2, 3, or 4.
     - For \( n = 2 \), \( m = 2 \) works since 2 is divisible by 1 and 2, but not by 3, 4, or 5.
     - For \( n = 3 \), \( m = 6 \) works since 6 is divisible by 1, 2, and 3, but not by 4, 5, or 6.

   - **Case 2: \( n > 3 \)**
     - If \( n + 1 \) is even, then \( n + 1 = 2^k \). For \( n + 2 \) and \( n + 3 \) to be powers of primes, they must be of the form \( p^k \). However, if \( 2^k < n + 3 < 2^{k + 1} \), it implies that \( n + 3 \) cannot be a power of 2, leading to a contradiction.
     - If \( n + 2 = 2^k \), then \( n + 1 \) and \( n + 3 \) must be analyzed similarly. If \( n + 3 = 3^m \), then we solve \( 3^m - 2^k = 1 \). The only solution to this equation is \( m = 2 \) and \( k = 3 \), giving \( n = 6 \).

5. **Verification:**
   - For \( n = 6 \), \( m = 60 \) works since 60 is divisible by 1, 2, 3, 4, 5, and 6, but not by 7, 8, or 9.

Thus, the natural numbers \( n \) that satisfy the given conditions are \( n = 1, 2, 6 \).

The final answer is \( \boxed{1, 2, 6} \).