Let $k, l, m, n$ be positive integers. Given that $k+l+m+n=km=ln$, find all possible values of $k+l+m+n$.
Given the equations \( k + l + m + n = km = ln \), we need to find all possible values of \( k + l + m + n \).

1. **Assume \( m \) is maximal:**
   Since \( m \) is the largest among \( k, l, m, n \), we have \( m \mid k + l + n \). Therefore, \( k + l + n \leq 3m \).

2. **Case 1: \( k + l + n = 3m \)**
   \[
   k + l + m + n = 3m + m = 4m
   \]
   Given \( k + l + m + n = km \), we have:
   \[
   4m = km \implies k = 4
   \]
   Similarly, \( k + l + m + n = ln \):
   \[
   4m = ln
   \]
   Since \( k = l = n = m = 4 \), we get:
   \[
   k + l + m + n = 4 + 4 + 4 + 4 = 16
   \]

3. **Case 2: \( k + l + n = 2m \)**
   \[
   k + l + m + n = 2m + m = 3m
   \]
   Given \( k + l + m + n = km \), we have:
   \[
   3m = km \implies k = 3
   \]
   Similarly, \( k + l + m + n = ln \):
   \[
   3m = ln
   \]
   Let \( l \) and \( n \) be the roots of the quadratic equation:
   \[
   x^2 - (2m-3)x + 3m = 0
   \]
   The discriminant \( D \) must be a perfect square:
   \[
   D = (2m-3)^2 - 4 \cdot 3m = 4m^2 - 12m + 9 - 12m = 4m^2 - 24m + 9
   \]
   \[
   D = (2m-6)^2 - 27
   \]
   For \( D \) to be a perfect square, we solve:
   \[
   (2m-6)^2 - 27 = k^2 \implies (2m-6)^2 - k^2 = 27
   \]
   This is a difference of squares:
   \[
   (2m-6-k)(2m-6+k) = 27
   \]
   Possible pairs \((a, b)\) that multiply to 27 are \((1, 27)\), \((3, 9)\), and \((-1, -27)\), \((-3, -9)\). Solving these pairs, we find:
   \[
   2m-6 = 9 \implies m = 7.5 \quad \text{(not an integer)}
   \]
   \[
   2m-6 = 15 \implies m = 10.5 \quad \text{(not an integer)}
   \]
   \[
   2m-6 = 3 \implies m = 4.5 \quad \text{(not an integer)}
   \]
   \[
   2m-6 = 6 \implies m = 6
   \]
   \[
   2m-6 = 12 \implies m = 9
   \]
   Therefore, valid values of \( m \) are 6 and 10:
   \[
   k + l + m + n = 3m = 18 \quad \text{and} \quad 30
   \]

4. **Case 3: \( k + l + n = m \)**
   \[
   k + l + m + n = m + m = 2m
   \]
   Given \( k + l + m + n = km \), we have:
   \[
   2m = km \implies k = 2
   \]
   Similarly, \( k + l + m + n = ln \):
   \[
   2m = ln
   \]
   Let \( l \) and \( n \) be the roots of the quadratic equation:
   \[
   x^2 - (m-2)x + 2m = 0
   \]
   The discriminant \( D \) must be a perfect square:
   \[
   D = (m-2)^2 - 4 \cdot 2m = m^2 - 4m + 4 - 8m = m^2 - 12m + 4
   \]
   \[
   D = (m-6)^2 - 32
   \]
   For \( D \) to be a perfect square, we solve:
   \[
   (m-6)^2 - 32 = k^2 \implies (m-6)^2 - k^2 = 32
   \]
   This is a difference of squares:
   \[
   (m-6-k)(m-6+k) = 32
   \]
   Possible pairs \((a, b)\) that multiply to 32 are \((1, 32)\), \((2, 16)\), \((4, 8)\), and their negatives. Solving these pairs, we find:
   \[
   m-6 = 8 \implies m = 14
   \]
   \[
   m-6 = 4 \implies m = 10
   \]
   Therefore, valid values of \( m \) are 12 and 15:
   \[
   k + l + m + n = 2m = 24 \quad \text{and} \quad 30
   \]

Combining all cases, the possible values of \( k + l + m + n \) are:
\[
16, 18, 24, 30
\]

The final answer is \(\boxed{16, 18, 24, 30}\).