Determine all pairs of positive integers $(m,n)$ such that m is but divisible by every integer from $1$ to $n$ (inclusive), but not divisible by $n + 1, n + 2$, and $n + 3$.
1. We start with the conditions given in the problem: \( m \) is divisible by every integer from \( 1 \) to \( n \) (inclusive), but not divisible by \( n + 1, n + 2 \), and \( n + 3 \). This can be written as:
   \[
   n! \mid m \quad \text{and} \quad d \nmid m \text{ for } d \in \{n+1, n+2, n+3\}
   \]

2. From the condition \( n! \mid m \), we know that \( m \) must be a multiple of \( n! \). However, \( m \) should not be divisible by \( n+1, n+2, \) and \( n+3 \). This implies:
   \[
   \gcd(d, n!) < d \quad \text{for} \quad d \in \{n+1, n+2, n+3\}
   \]

3. We need to analyze the greatest common divisor (gcd) conditions. Notice that for any non-prime \( k > 4 \), we have:
   \[
   \gcd((k-1)!, k) = k, \quad \gcd((k-1)!, k+1) = k+1, \quad \gcd((k-1)!, k+2) = k+2
   \]
   This is because if \( k \) is non-prime and greater than 4, then \( k \), \( k+1 \), and \( k+2 \) will have prime factors that are less than or equal to \( k \), and thus will divide \((k-1)!\).

4. Therefore, for \( n+1, n+2, n+3 \) to not divide \( m \), they must be prime numbers. This is because if any of \( n+1, n+2, n+3 \) were composite, their prime factors would be less than or equal to \( n \), and thus would divide \( n! \).

5. However, it is impossible for three consecutive numbers \( n+1, n+2, n+3 \) to all be prime. Therefore, the only possible values for \( n \) are those for which \( n+1, n+2, n+3 \) do not all need to be prime.

6. Let's check small values of \( n \):
   - For \( n = 1 \):
     \[
     n+1 = 2, \quad n+2 = 3, \quad n+3 = 4
     \]
     Here, \( 2 \) and \( 3 \) are primes, and \( 4 \) is not. So, \( n = 1 \) is a valid solution. This implies any \( (m, 1) \) such that \( \gcd(m, 6) = 1 \) works because \( 6 = 1! \cdot 6 \).

   - For \( n = 2 \):
     \[
     n+1 = 3, \quad n+2 = 4, \quad n+3 = 5
     \]
     Here, \( 3 \) and \( 5 \) are primes, and \( 4 \) is not. So, \( n = 2 \) is a valid solution. This implies any \( (m, 2) \) such that \( \gcd(m, 12) = 1 \) works because \( 12 = 2! \cdot 6 \).

7. For \( n \geq 3 \), it is impossible for \( n+1, n+2, n+3 \) to all be primes, as shown in step 4.

Conclusion:
The only possible values for \( n \) are \( n = 1 \) and \( n = 2 \).

The final answer is \( \boxed{ (m, 1) } \) such that \(\gcd(m, 6) = 1\) and \((m, 2)\) such that \(\gcd(m, 12) = 1\).