Let $m$ be a positive integer and $a$ and $b$ be distinct positive integers strictly greater than $m^2$ and strictly less than $m^2+m$. Find all integers $d$ such that $m^2 < d < m^2+m$ and $d$ divides $ab$.
1. **Identify the range for \(a\) and \(b\):**
   Given that \(a\) and \(b\) are distinct positive integers such that \(m^2 < a, b < m^2 + m\), we can write:
   \[
   m^2 < a < m^2 + m \quad \text{and} \quad m^2 < b < m^2 + m
   \]

2. **Identify the range for \(d\):**
   We need to find all integers \(d\) such that:
   \[
   m^2 < d < m^2 + m \quad \text{and} \quad d \mid ab
   \]

3. **Consider \(d = a\) and \(d = b\):**
   Clearly, \(d = a\) and \(d = b\) satisfy the conditions since \(a\) and \(b\) are within the specified range and both divide \(ab\).

4. **Consider \(d \neq a\) and \(d \neq b\):**
   Suppose \(d\) is an integer such that \(d \neq a\) and \(d \neq b\) but \(d \mid ab\). This implies there exists some positive integer \(k\) such that:
   \[
   ab = dk
   \]

5. **Factorization and properties of \(d\):**
   It is well known that if \(d \mid ab\), then \(d\) can be expressed in terms of the prime factors of \(a\) and \(b\). Let:
   \[
   a = pq \quad \text{and} \quad b = rs
   \]
   where \(p, q, r, s\) are integers and \(d = pr\) and \(k = qs\) with \(\gcd(q, s) = 1\).

6. **Range constraints:**
   Since \(m^2 < a, b < m^2 + m\), we have:
   \[
   m^2 < pq < m^2 + m \quad \text{and} \quad m^2 < rs < m^2 + m
   \]
   and since \(d = pr\), we need:
   \[
   m^2 < pr < m^2 + m
   \]

7. **Contradiction:**
   Given \(a = pq\) and \(b = rs\), and considering \(d = pr\), we need to check if \(d\) can be within the range \(m^2 < d < m^2 + m\). However, since \(p\) and \(r\) are both less than \(m\) (as \(p, r < m\)), we get:
   \[
   pr < m^2
   \]
   which contradicts the condition \(m^2 < pr\).

8. **Conclusion:**
   Since any \(d\) other than \(a\) or \(b\) would lead to a contradiction, the only possible values for \(d\) are \(a\) and \(b\).

The final answer is \( \boxed{ d = a } \) and \(d = b\).