Find all positive integers $n$ with the  folowing property: for all triples ($a$,$b$,$c$) of positive real there is a triple of non negative integers ($l$,$j$,$k$) such that $an^k$, $bn^j$ and $cn^l$ are sides of a non degenate triangle
To solve the problem, we need to find all positive integers \( n \) such that for any positive real numbers \( a, b, c \), there exist non-negative integers \( l, j, k \) such that \( an^k, bn^j, \) and \( cn^l \) can form the sides of a non-degenerate triangle. 

A non-degenerate triangle must satisfy the triangle inequality:
1. \( an^k + bn^j > cn^l \)
2. \( an^k + cn^l > bn^j \)
3. \( bn^j + cn^l > an^k \)

We will analyze the problem for different values of \( n \).

1. **Case \( n = 1 \):**
   - If \( n = 1 \), then \( an^k = a, bn^j = b, \) and \( cn^l = c \).
   - The inequalities become \( a + b > c \), \( a + c > b \), and \( b + c > a \).
   - These inequalities must hold for any positive real numbers \( a, b, c \). However, this is not always true. For example, if \( a = 1, b = 2, c = 3 \), then \( 1 + 2 \not> 3 \). Therefore, \( n = 1 \) does not satisfy the condition.

2. **Case \( n \geq 5 \):**
   - Let us consider \( (a, b, c) = (1, 2, 3) \).
   - We need to check if there exist non-negative integers \( l, j, k \) such that \( n^k, 2n^j, 3n^l \) can form a non-degenerate triangle.
   - For \( n \geq 5 \), we have:
     - \( n^k + 2n^j > 3n^l \)
     - \( n^k + 3n^l > 2n^j \)
     - \( 2n^j + 3n^l > n^k \)
   - If we choose \( k = 0, j = 0, l = 0 \), then the inequalities become:
     - \( 1 + 2 > 3 \) which is false.
   - Therefore, \( n \geq 5 \) does not satisfy the condition.

3. **Case \( n = 2, 3, 4 \):**
   - We need to show that for any positive real numbers \( a, b, c \), there exist non-negative integers \( l, j, k \) such that \( an^k, bn^j, cn^l \) can form a non-degenerate triangle.
   - Without loss of generality, let \( a = 1 \) and \( 1 \leq b \leq c < n \).
   - We need to check the following cases:
     - If \( c < b + 1 \), then \( (l, j, k) = (0, 0, 0) \) works because \( 1, b, c \) form a non-degenerate triangle.
     - If \( b + c > n \), then \( (l, j, k) = (1, 0, 0) \) works because \( n, b, c \) form a non-degenerate triangle.
     - If \( n + c > nb \), then \( (l, j, k) = (1, 1, 0) \) works because \( n, nb, c \) form a non-degenerate triangle.
     - If \( c \geq b + 1 \), \( b + c \leq n \), and \( n + c \leq nb \):
       - We have \( \frac{n+1}{2}c \geq \frac{n+1}{2}b + \frac{n+1}{2} \) and \( nb \geq n + c \).
       - This implies \( \frac{n-1}{2}(b+c) \geq \frac{3n+1}{2} \).
       - Simplifying, we get \( \frac{n(n-1)}{2} \geq \frac{3n+1}{2} \).
       - This inequality holds for \( n = 2, 3, 4 \), but not for \( n \geq 5 \).

Therefore, the only values of \( n \) that satisfy the given condition are \( n = 2, 3, 4 \).

The final answer is \( \boxed{2, 3, 4} \).