Find all positive integers $k$ for which the equation: $$ \text{lcm}(m,n)-\text{gcd}(m,n)=k(m-n)$$ has no solution in integers positive $(m,n)$ with $m\neq n$.

1. **Substitute \( m = dx \) and \( n = dy \) with \(\gcd(x, y) = 1\):**

   Given the equation:
   \[
   \text{lcm}(m, n) - \text{gcd}(m, n) = k(m - n)
   \]
   Substitute \( m = dx \) and \( n = dy \):
   \[
   \text{lcm}(dx, dy) - \text{gcd}(dx, dy) = k(dx - dy)
   \]
   Since \(\gcd(dx, dy) = d\) and \(\text{lcm}(dx, dy) = dxy\), the equation becomes:
   \[
   dxy - d = k(dx - dy)
   \]

2. **Simplify the equation:**

   Factor out \( d \) from both sides:
   \[
   d(xy - 1) = kd(x - y)
   \]
   Divide both sides by \( d \) (since \( d \neq 0 \)):
   \[
   xy - 1 = k(x - y)
   \]

3. **Rearrange the equation:**

   Rearrange to isolate terms involving \( x \) and \( y \):
   \[
   xy - kx + ky = 1
   \]
   \[
   x(y - k) = 1 - ky
   \]

4. **Analyze the equation for \( k = 2 \):**

   Assume \( k = 2 \):
   \[
   x(y - 2) = 1 - 2y
   \]
   Since \( x \) and \( y \) are positive integers and \( x, y \geq 1 \), the right-hand side \( 1 - 2y \) must be negative for \( y \geq 2 \). This implies \( y = 1 \):
   \[
   x(1 - 2) = 1 - 2 \cdot 1
   \]
   \[
   -x = -1 \implies x = 1
   \]
   This gives \( x = y = 1 \), which contradicts \( x \neq y \). Therefore, there is no solution for \( k = 2 \).

5. **Construct solutions for \( k = 1 \) and \( k > 2 \):**

   - For \( k = 1 \):
     \[
     x(y - 1) = 1 - y
     \]
     Take \( m = 6 \) and \( n = 3 \):
     \[
     \text{lcm}(6, 3) - \text{gcd}(6, 3) = 6 - 3 = 3 = 1(6 - 3)
     \]
     This is a valid solution.

   - For \( k > 2 \):
     Take \( (x, y) = (k^2 - k - 1, k - 1) \):
     \[
     x(y - k) = (k^2 - k - 1)(k - 1 - k) = (k^2 - k - 1)(-1) = -k^2 + k + 1
     \]
     \[
     1 - ky = 1 - k(k - 1) = 1 - k^2 + k
     \]
     Both sides are equal, and by the Euclidean Algorithm, \(\gcd(k^2 - k - 1, k - 1) = 1\), confirming that this pair is a solution for all \( k > 2 \).

6. **Conclusion:**

   The only \( k \) for which the equation has no solution in positive integers \( (m, n) \) with \( m \neq n \) is \( k = 2 \).

The final answer is \( \boxed{ k = 2 } \).