Let $m$ be a given positive integer which has a prime divisor greater than $\sqrt {2m} +1 $. Find the minimal positive integer $n$ such that there exists a finite set $S$ of distinct positive integers satisfying the following two conditions:

[b]I.[/b] $m\leq x\leq n$ for all $x\in S$;

[b]II.[/b] the product of all elements in $S$ is the square of an integer.
1. **Restate the problem with correct translation:**
   We need to find the minimal positive integer \( n \) such that there exists a finite set \( S \) of distinct positive integers satisfying the following two conditions:
   - The smallest element of \( S \) is \( m \), and the largest element of \( S \) is \( n \).
   - The product of all elements in \( S \) is the square of an integer.

2. **Given condition:**
   \( m \) is a positive integer with a prime divisor \( p \) such that \( p > \sqrt{2m} + 1 \).

3. **Determine the minimal \( n \):**
   The answer is \( \boxed{n = m + p} \), where \( p \) is the largest prime factor of \( m \). By the given condition, \( p > \sqrt{2m} + 1 \). Clearly, \( n \) must be at least \( m + p \) because \( m < p^2 \) implies \( p \) divides \( m \) only once, so we need another factor of \( p \).

4. **Construct the corresponding set \( S \) for \( n = m + p \):**
   Write \( m = kp \) (so that \( n = (k+1)p \)). Then, \( p > \sqrt{2kp} + 1 \) implies:
   \[
   (p-1)^2 > 2kp \implies (p-1)^2 \geq 2kp + 1 \implies p \geq 2k + 2 > 2k + 1
   \]
   Clearly, \( p \) is odd. We have two cases:

5. **Case I: \( p = 4k + 1 \) or \( p = 4k + 3 \):**
   In this case, notice that \( kp < 4k(k+1) < (k+1)p \). So if we take \( S = \{kp, 4k(k+1), (k+1)p\} \), then the product of elements of \( S \) is:
   \[
   kp \cdot 4k(k+1) \cdot (k+1)p = 4p^2k^2(k+1)^2
   \]
   which is a perfect square.

6. **Case II: \( p \neq 4k + 1 \) and \( p \neq 4k + 3 \):**
   In this case, we take:
   \[
   S = \left\{kp, k(p+1), \frac{(2k+1)(p-1)}{2}, \frac{(2k+1)(p+1)}{2}, (k+1)(p-1), (k+1)p\right\}
   \]
   The product of elements of \( S \) is:
   \[
   kp \cdot k(p+1) \cdot \frac{(2k+1)(p-1)}{2} \cdot \frac{(2k+1)(p+1)}{2} \cdot (k+1)(p-1) \cdot (k+1)p = \frac{p^2(p-1)^2(p+1)^2k^2(k+1)^2(2k+1)^2}{4}
   \]
   which is a perfect square. It is enough to verify that the 6 elements are distinct and in the required range.

7. **Verify inequalities:**
   We have the trivial inequalities:
   \[
   kp < k(p+1), \quad (k+1)(p-1) < (k+1)p, \quad \frac{(2k+1)(p-1)}{2} < (k+1)(p-1), \quad k(p+1) < \frac{(2k+1)(p+1)}{2}, \quad \frac{(2k+1)(p-1)}{2} < \frac{(2k+1)(p+1)}{2}
   \]
   Further, we can verify that the following inequalities are equivalent to \( p > 2k + 1 \):
   \[
   kp < \frac{(2k+1)(p-1)}{2}, \quad \frac{(2k+1)(p+1)}{2} < (k+1)p, \quad k(p+1) < (k+1)(p-1)
   \]
   Therefore, the only discrepancy occurs when:
   \[
   \frac{(2k+1)(p-1)}{2} = k(p+1) \quad \text{or} \quad \frac{(2k+1)(p+1)}{2} = (k+1)(p-1)
   \]
   In the first equality, we get \( p = 4k + 1 \), and in the second, we get \( p = 4k + 3 \), both of which we have excluded. 

\(\blacksquare\)