Find the positive integers $n$ with $n \geq 4$ such that $[\sqrt{n}]+1$ divides $n-1$ and $[\sqrt{n}]-1$ divides $n+1$.
[hide="Remark"]This problem can be solved in a similar way with the one given at [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=97&year=2006]Cono Sur Olympiad 2006[/url], problem 5.[/hide]
1. Let \( a = \lfloor \sqrt{n} \rfloor \). Then we can write \( n \) as:
   \[
   n = a^2 + r \quad \text{where} \quad 0 \leq r < 2a + 1
   \]

2. Given that \( a+1 \) divides \( n-1 \), we can express this as:
   \[
   a+1 \mid n-1
   \]
   Substituting \( n = a^2 + r \) into the equation, we get:
   \[
   a+1 \mid (a^2 + r - 1)
   \]
   Since \( a^2 - 1 = (a-1)(a+1) \), we have:
   \[
   a+1 \mid (a^2 - 1 + r) \implies a+1 \mid r
   \]
   Given \( 0 \leq r < 2a + 1 \), the possible values for \( r \) are \( 0 \) or \( a+1 \).

3. Next, given that \( a-1 \) divides \( n+1 \), we can express this as:
   \[
   a-1 \mid n+1
   \]
   Substituting \( n = a^2 + r \) into the equation, we get:
   \[
   a-1 \mid (a^2 + r + 1)
   \]
   Since \( a^2 + 1 = (a-1)(a+1) + 2 \), we have:
   \[
   a-1 \mid (a^2 + r + 1) \implies a-1 \mid (r + 2)
   \]

4. Consider the case \( r = 0 \):
   \[
   n = a^2
   \]
   Substituting \( r = 0 \) into the divisibility condition \( a-1 \mid 2 \), we get:
   \[
   a-1 \mid 2
   \]
   The possible values for \( a-1 \) are \( 1 \) and \( 2 \), giving \( a = 2 \) or \( a = 3 \). Thus, the possible values for \( n \) are:
   \[
   n = 2^2 = 4 \quad \text{or} \quad n = 3^2 = 9
   \]

5. Consider the case \( r = a+1 \):
   \[
   n = a^2 + a + 1
   \]
   Substituting \( r = a+1 \) into the divisibility condition \( a-1 \mid (a+1 + 2) \), we get:
   \[
   a-1 \mid (a + 3)
   \]
   Simplifying, we have:
   \[
   a-1 \mid 4
   \]
   The possible values for \( a-1 \) are \( 1, 2, \) and \( 4 \), giving \( a = 2, 3, \) and \( 5 \). Thus, the possible values for \( n \) are:
   \[
   n = 2^2 + 2 + 1 = 7, \quad n = 3^2 + 3 + 1 = 13, \quad n = 5^2 + 5 + 1 = 31
   \]

Combining all the cases, the positive integers \( n \) with \( n \geq 4 \) that satisfy the given conditions are:
\[
n = 4, 7, 9, 13, 31
\]

The final answer is \( \boxed{4, 7, 9, 13, 31} \).