Find all pairs of positive integers $(m,n)$ such that 
$$lcm(1,2,\dots,n)=m!$$
where $lcm(1,2,\dots,n)$ is the smallest positive integer multiple of all $1,2,\dots n-1$ and $n$.
1. **Understanding the problem**: We need to find all pairs of positive integers \((m, n)\) such that the least common multiple (LCM) of the first \(n\) positive integers is equal to \(m!\).

2. **Using Bertrand's Postulate (Chebyshev's Theorem)**: Bertrand's Postulate states that for any integer \(n > 1\), there is always at least one prime \(p\) such that \(n < p < 2n\). This implies that the largest prime \(p\) not greater than \(n\) is at least \(0.5n\).

3. **Analyzing the LCM and factorial**: 
   - The LCM of the first \(n\) positive integers, \( \text{lcm}(1, 2, \ldots, n) \), is the smallest number that is divisible by each of the integers from 1 to \(n\).
   - The factorial \(m!\) is the product of all positive integers up to \(m\).

4. **Prime factorization and valuation**:
   - For any prime \(p\), the \(p\)-adic valuation \(v_p(k)\) of an integer \(k\) is the highest power of \(p\) that divides \(k\).
   - The \(p\)-adic valuation of \(m!\) is given by:
     \[
     v_p(m!) = \left\lfloor \frac{m}{p} \right\rfloor + \left\lfloor \frac{m}{p^2} \right\rfloor + \left\lfloor \frac{m}{p^3} \right\rfloor + \cdots
     \]
   - The \(p\)-adic valuation of \(\text{lcm}(1, 2, \ldots, n)\) is:
     \[
     v_p(\text{lcm}(1, 2, \ldots, n)) = \max(v_p(1), v_p(2), \ldots, v_p(n))
     \]

5. **Considering the power of 2**:
   - The \(2\)-adic valuation of \(m!\) is at least \(\frac{m}{2}\) because:
     \[
     v_2(m!) = \left\lfloor \frac{m}{2} \right\rfloor + \left\lfloor \frac{m}{4} \right\rfloor + \left\lfloor \frac{m}{8} \right\rfloor + \cdots \geq \frac{m}{2}
     \]
   - The \(2\)-adic valuation of \(\text{lcm}(1, 2, \ldots, n)\) is \(\log_2(n)\) because the highest power of 2 that divides any number up to \(n\) is \(2^{\lfloor \log_2(n) \rfloor}\).

6. **Inequality analysis**:
   - We have:
     \[
     v_2(m!) \geq \frac{m}{2} \quad \text{and} \quad v_2(\text{lcm}(1, 2, \ldots, n)) = \log_2(n)
     \]
   - Therefore:
     \[
     \log_2(n) \geq \frac{m}{2}
     \]
   - Since \(m \geq 0.5n\), we get:
     \[
     \log_2(n) \geq 0.25n
     \]

7. **Solving the inequality**:
   - The inequality \(\log_2(n) \geq 0.25n\) implies that \(n\) must be relatively small. By solving this inequality, we find that \(n \leq 16\).

8. **Checking values of \(n\) from 1 to 16**:
   - For each \(n\) from 1 to 16, we check if \(\text{lcm}(1, 2, \ldots, n) = m!\) for some \(m\).
   - After checking, we find that the pairs \((m, n)\) that satisfy the condition are:
     \[
     (1, 1), (2, 2), (3, 3)
     \]

The final answer is \( \boxed{ (1, 1), (2, 2), (3, 3) } \)