Let $M \geq 1$ be a real number. Determine all natural numbers $n$ for which there exist distinct natural numbers $a$, $b$, $c > M$, such that
$n = (a,b) \cdot (b,c) + (b,c) \cdot (c,a) + (c,a) \cdot (a,b)$
(where $(x,y)$ denotes the greatest common divisor of natural numbers $x$ and $y$).
1. **Define the Problem and Key Notations:**
   Let \( M \geq 1 \) be a real number. We need to determine all natural numbers \( n \) for which there exist distinct natural numbers \( a, b, c > M \) such that:
   \[
   n = \gcd(a,b) \cdot \gcd(b,c) + \gcd(b,c) \cdot \gcd(c,a) + \gcd(c,a) \cdot \gcd(a,b)
   \]
   where \(\gcd(x,y)\) denotes the greatest common divisor of natural numbers \( x \) and \( y \).

2. **Introduce the Concept of "Good" Numbers:**
   Call a positive integer \( n \) *good* if it satisfies the given condition. We claim that the good positive integers \( n \) are those for which \( v_2(n) \) (the 2-adic valuation of \( n \)) is even and \( n \) is not a power of 4 (including 1).

3. **Claim 1: \( n \) is good if and only if \( 4n \) is good.**
   - **Proof:**
     - If \( n \) is good, then \( 4n \) is also good by taking \( (2a, 2b, 2c) \) instead of \( (a, b, c) \). This is because:
       \[
       \gcd(2a, 2b) = 2\gcd(a, b), \quad \gcd(2b, 2c) = 2\gcd(b, c), \quad \gcd(2c, 2a) = 2\gcd(c, a)
       \]
       Thus,
       \[
       4n = 4(\gcd(a,b) \cdot \gcd(b,c) + \gcd(b,c) \cdot \gcd(c,a) + \gcd(c,a) \cdot \gcd(a,b))
       \]
     - Conversely, if \( 4n \) is good, then we claim that \( a, b, c \) must all be even. If \( a, b, c \) are all odd, then \( K \) is odd, which is a contradiction. If \( a \) is odd and \( b, c \) are even, then all \(\gcd\)'s are odd, making \( K \) odd, which is also a contradiction. Similarly, if \( a, b \) are even and \( c \) is odd, \( K \) is odd, which is again a contradiction. Thus, \( a, b, c \) must all be even.

4. **Claim 2: If \( n \equiv 2 \pmod{4} \), then \( n \) is not good.**
   - **Proof:**
     - Assume \( n \equiv 2 \pmod{4} \) is good. Then, by the same reasoning as in Claim 1, \( a, b, c \) must all be even, making \( n \) a multiple of 4, which is a contradiction.

5. **Analysis of \( v_2(n) \):**
   - If \( v_2(n) \) is odd, then if \( v_2(n) = 1 \), we have a contradiction by Claim 1. If \( v_2(n) \geq 3 \), using Claim 1, we may assume \( v_2(n) = 1 \), leading to a contradiction again.
   - If \( n = 4 \) is good, then by Claim 1, \( n = 1 \) is good, which is a contradiction.

6. **Construction for \( v_2(n) \) even and \( n \) not a power of 4:**
   - If \( v_2(n) = 0 \) (i.e., \( n \) is odd), we can use the construction:
     \[
     a = nk, \quad b = n(k+1), \quad c = p
     \]
     with \( k > M \) and \( p \) a prime such that \( p > n(k+1) \). Then:
     \[
     \gcd(a, c) = \gcd(b, c) = 1, \quad \gcd(a, b) = n
     \]
     Thus,
     \[
     \gcd(a, b) \cdot \gcd(b, c) + \gcd(b, c) \cdot \gcd(c, a) + \gcd(c, a) \cdot \gcd(a, b) = n + 1 + n = 2n + 1
     \]

7. **Conclusion:**
   - The good numbers are those for which \( v_2(n) \) is even and \( n \) is not a power of 4 (including 1).

The final answer is \( \boxed{ n } \) such that \( v_2(n) \) is even and \( n \) is not a power of 4 (including 1).