For every positive integer $N$, let $\sigma(N)$ denote the sum of the positive integer divisors of $N$. Find all integers $m\geq n\geq 2$ satisfying \[\frac{\sigma(m)-1}{m-1}=\frac{\sigma(n)-1}{n-1}=\frac{\sigma(mn)-1}{mn-1}.\]

[i]Ankan Bhattacharya[/i]
To solve the problem, we need to find all integers \( m \geq n \geq 2 \) such that:
\[
\frac{\sigma(m)-1}{m-1} = \frac{\sigma(n)-1}{n-1} = \frac{\sigma(mn)-1}{mn-1}.
\]

1. **Define the common ratio:**
   Let \( K \) be the common value of the given fractions:
   \[
   \frac{\sigma(m)-1}{m-1} = \frac{\sigma(n)-1}{n-1} = \frac{\sigma(mn)-1}{mn-1} = K.
   \]

2. **Express \(\sigma\) in terms of \(K\):**
   From the definition of \( K \), we have:
   \[
   \sigma(m) = K(m-1) + 1,
   \]
   \[
   \sigma(n) = K(n-1) + 1,
   \]
   \[
   \sigma(mn) = K(mn-1) + 1.
   \]

3. **Substitute \(\sigma(m)\) and \(\sigma(n)\) into \(\sigma(mn)\):**
   Using the multiplicative property of the divisor sum function \(\sigma\) for coprime integers, we have:
   \[
   \sigma(mn) = \sigma(m) \sigma(n).
   \]
   Substituting the expressions for \(\sigma(m)\) and \(\sigma(n)\):
   \[
   \sigma(mn) = [K(m-1) + 1][K(n-1) + 1].
   \]

4. **Expand and simplify:**
   \[
   \sigma(mn) = K^2(m-1)(n-1) + K(m-1) + K(n-1) + 1.
   \]
   Equate this to the expression for \(\sigma(mn)\) from step 2:
   \[
   K(mn-1) + 1 = K^2(m-1)(n-1) + K(m-1) + K(n-1) + 1.
   \]

5. **Cancel the constant terms and simplify:**
   \[
   K(mn-1) = K^2(m-1)(n-1) + K(m-1) + K(n-1).
   \]
   Divide both sides by \( K \):
   \[
   mn - 1 = K(m-1)(n-1) + (m-1) + (n-1).
   \]
   \[
   mn - 1 = K(m-1)(n-1) + m + n - 2.
   \]

6. **Rearrange the equation:**
   \[
   mn - m - n + 1 = K(m-1)(n-1).
   \]
   \[
   (m-1)(n-1) = K(m-1)(n-1).
   \]

7. **Solve for \(K\):**
   If \( m \neq 1 \) and \( n \neq 1 \), we can divide both sides by \((m-1)(n-1)\):
   \[
   1 = K.
   \]

8. **Verify the solution:**
   If \( K = 1 \), then:
   \[
   \sigma(m) = m,
   \]
   \[
   \sigma(n) = n,
   \]
   \[
   \sigma(mn) = mn.
   \]
   This implies that \( m \) and \( n \) must be powers of the same prime \( p \). Let \( m = p^e \) and \( n = p^f \) with \( e \geq f \).

9. **Conclusion:**
   The only solutions are of the form \( (m, n) = (p^e, p^f) \) for some prime \( p \) and integers \( e \geq f \geq 1 \).

The final answer is \( \boxed{ (m, n) = (p^e, p^f) } \) for some prime \( p \) and integers \( e \geq f \geq 1 \).