Find all functions $f : \mathbb{N} \rightarrow \mathbb{N} $ satisfying
\[ f(mn) = \operatorname{lcm} (m,n) \cdot \gcd( f(m), f(n) ) \]
for all positive integer $m,n$.
1. **Initial Setup and Notation:**
   Let \( f(1) = k \). We denote the given functional equation as \( P(m, n) \):
   \[
   P(m, n) \implies f(mn) = \operatorname{lcm}(m, n) \cdot \gcd(f(m), f(n))
   \]

2. **Applying \( P(m, 1) \):**
   \[
   P(m, 1) \implies f(m) = \operatorname{lcm}(m, 1) \cdot \gcd(f(m), f(1))
   \]
   Since \(\operatorname{lcm}(m, 1) = m\) and \(f(1) = k\), we have:
   \[
   f(m) = m \cdot \gcd(f(m), k)
   \]

3. **Applying \( P(km, 1) \):**
   \[
   P(km, 1) \implies f(km) = \operatorname{lcm}(km, 1) \cdot \gcd(f(km), k)
   \]
   Since \(\operatorname{lcm}(km, 1) = km\), we have:
   \[
   f(km) = km \cdot \gcd(f(km), k)
   \]
   Assume \( f(km) = k^2m \). This assumption will be verified later.

4. **Applying \( P(m, kn) \):**
   \[
   P(m, kn) \implies f(mkn) = \operatorname{lcm}(m, kn) \cdot \gcd(f(m), f(kn))
   \]
   Since \(\operatorname{lcm}(m, kn) = \frac{mkn}{\gcd(m, kn)}\), we have:
   \[
   f(mkn) = \frac{mkn}{\gcd(m, kn)} \cdot \gcd(f(m), f(kn))
   \]
   Using the assumption \( f(kn) = k^2n \), we get:
   \[
   f(mkn) = \frac{mkn}{\gcd(m, kn)} \cdot \gcd(f(m), k^2n)
   \]

5. **Simplifying the GCD:**
   \[
   \gcd(f(m), k^2n) = k \cdot \gcd(f(m), kn)
   \]
   Therefore:
   \[
   f(mkn) = \frac{mkn}{\gcd(m, kn)} \cdot k \cdot \gcd(f(m), kn)
   \]

6. **Conclusion:**
   From the above steps, we see that \( k \) must divide \( f(m) \) for all \( m \). Hence, we can write \( f(m) = km \) for all \( m \).

7. **Verification:**
   We check that \( f(m) = km \) satisfies the original functional equation:
   \[
   f(mn) = k(mn) = kmn
   \]
   \[
   \operatorname{lcm}(m, n) \cdot \gcd(f(m), f(n)) = \operatorname{lcm}(m, n) \cdot \gcd(km, kn) = \operatorname{lcm}(m, n) \cdot k \cdot \gcd(m, n)
   \]
   Since \(\operatorname{lcm}(m, n) \cdot \gcd(m, n) = mn\), we have:
   \[
   \operatorname{lcm}(m, n) \cdot k \cdot \gcd(m, n) = kmn
   \]
   Thus, \( f(mn) = \operatorname{lcm}(m, n) \cdot \gcd(f(m), f(n)) \) holds true.

The final answer is \( \boxed{ f(x) = kx } \) for some \( k \in \mathbb{N} \).