Determine all natural numbers $n$ of the form $n=[a,b]+[b,c]+[c,a]$ where $a,b,c$ are positive integers and $[u,v]$ is the least common multiple of the integers $u$ and $v$.
1. **Substitute the given expressions:**
   Let \( a = dpqx \), \( b = dpry \), and \( c = dqrz \), where \( \gcd(a, b, c) = d \), and \( p, q, r \) are the common factors of \( a, b \), \( a, c \), and \( b, c \) respectively, besides \( d \). Also, \( x, y, z \) are the lone factors of \( a, b, c \) respectively.

2. **Express the least common multiples:**
   We need to find the least common multiples \([a, b]\), \([b, c]\), and \([c, a]\):
   \[
   [a, b] = \frac{ab}{\gcd(a, b)}, \quad [b, c] = \frac{bc}{\gcd(b, c)}, \quad [c, a] = \frac{ca}{\gcd(c, a)}
   \]

3. **Calculate the least common multiples:**
   \[
   \gcd(a, b) = dpr, \quad \gcd(b, c) = dqr, \quad \gcd(c, a) = dpq
   \]
   Therefore,
   \[
   [a, b] = \frac{dpqx \cdot dpry}{dpr} = dpqxy, \quad [b, c] = \frac{dpry \cdot dqrz}{dqr} = dpryz, \quad [c, a] = \frac{dqrz \cdot dpqx}{dpq} = dqrxz
   \]

4. **Sum the least common multiples:**
   \[
   n = [a, b] + [b, c] + [c, a] = dpqxy + dpryz + dqrxz
   \]
   Factor out the common term \( dpqr \):
   \[
   n = dpqr(xy + yz + xz)
   \]
   Let \( M = dpqr \), then:
   \[
   n = M(xy + yz + xz)
   \]

5. **Analyze the expression \( xy + yz + xz \):**
   Since \( x, y, z \) are coprime in pairs, \( xy + yz + xz \) is an odd number. This is because the sum of products of coprime numbers is odd.

6. **Conclusion about \( n \):**
   Since \( xy + yz + xz \) is odd, \( n \) cannot be a power of 2. Additionally, by substituting \( y = z = 1 \), we get:
   \[
   n = M(2x + 1)
   \]
   This shows that any number different from a power of 2 can be written in this form.

The final answer is \( \boxed{ n = M(xy + yz + xz) } \)