Determine all integer $n > 1$ such that
\[\gcd \left( n, \dfrac{n-m}{\gcd(n,m)} \right) = 1\]
for all integer $1 \le m < n$.
To determine all integers \( n > 1 \) such that
\[
\gcd \left( n, \dfrac{n-m}{\gcd(n,m)} \right) = 1
\]
for all integers \( 1 \le m < n \), we will analyze the problem step by step.

1. **Consider a prime power \( p^k \| n \)**:
   - Let \( p \) be the smallest prime divisor of \( n \).
   - We need to check when \( p \) divides \( \frac{n-m}{\gcd(n,m)} \).

2. **Analyze the condition \( p \mid \frac{n-m}{\gcd(n,m)} \)**:
   - This condition is equivalent to \( v_p(n-m) > v_p(\gcd(n,m)) \), where \( v_p(x) \) denotes the highest power of \( p \) dividing \( x \).

3. **Case when \( p \mid m \)**:
   - Let \( v_p(m) = a \).
   - If \( a < k \), then \( v_p(n-m) = v_p(\gcd(n,m)) = a \), so \( p \) does not divide the fraction \( \frac{n-m}{\gcd(n,m)} \).
   - If \( a = k \), then \( v_p(n-m) \ge k+1 \).

4. **Existence of \( m \) such that \( v_p(m) = k \)**:
   - If \( n \) is not a prime power of \( p \), there exists such an \( m \) with \( v_p(m) = k \) and \( m < n \).
   - For example, take \( m \) such that \( m/p^k \equiv n/p^k \pmod{p} \) and \( m < p^{k+1} < n \).

5. **Contradiction for non-prime power \( n \)**:
   - If \( n \) is not a prime power of \( p \), then \( p \mid \frac{n-m}{\gcd(n,m)} \), leading to \( \gcd(n, \frac{n-m}{\gcd(n,m)}) \ge p \), which contradicts the given condition.

6. **Conclusion for \( n = p^k \)**:
   - When \( n = p^k \), the problem holds because \( v_p(n-m) = v_p(m) \) and \( v_p(\gcd(n,m)) = v_p(m) \), ensuring the gcd is always 1.

Thus, the only integers \( n > 1 \) that satisfy the given condition for all \( 1 \le m < n \) are the prime powers.

The final answer is \( \boxed{ n = p^k } \) where \( p \) is a prime and \( k \) is a positive integer.