For some positive integer $n$, the sum of all odd positive integers between $n^2-n$ and $n^2+n$ is a number between $9000$ and $10000$, inclusive. Compute $n$.

[i]2020 CCA Math Bonanza Lightning Round #3.1[/i]
1. First, observe that \( n^2 - n \) and \( n^2 + n \) are both even numbers. We need to find the sum of all odd integers between these two even numbers.

2. The sequence of odd integers between \( n^2 - n \) and \( n^2 + n \) is:
   \[
   n^2 - n + 1, \, n^2 - n + 3, \, \ldots, \, n^2 + n - 1
   \]

3. This sequence is an arithmetic sequence with the first term \( a = n^2 - n + 1 \) and the common difference \( d = 2 \).

4. The number of terms in this sequence can be found by solving for \( k \) in the equation:
   \[
   n^2 - n + 1 + 2(k-1) = n^2 + n - 1
   \]
   Simplifying, we get:
   \[
   n^2 - n + 1 + 2k - 2 = n^2 + n - 1
   \]
   \[
   2k - 1 = 2n - 1
   \]
   \[
   2k = 2n
   \]
   \[
   k = n
   \]
   So, there are \( n \) terms in the sequence.

5. The sum of an arithmetic sequence is given by:
   \[
   S = \frac{k}{2} \left( a + l \right)
   \]
   where \( k \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Here, \( k = n \), \( a = n^2 - n + 1 \), and \( l = n^2 + n - 1 \).

6. Substituting these values, we get:
   \[
   S = \frac{n}{2} \left( (n^2 - n + 1) + (n^2 + n - 1) \right)
   \]
   Simplifying inside the parentheses:
   \[
   S = \frac{n}{2} \left( 2n^2 \right)
   \]
   \[
   S = \frac{n}{2} \cdot 2n^2
   \]
   \[
   S = n^3
   \]

7. We are given that the sum \( S \) is between 9000 and 10000, inclusive:
   \[
   9000 \leq n^3 \leq 10000
   \]

8. To find \( n \), we take the cube root of the bounds:
   \[
   \sqrt[3]{9000} \approx 20.88
   \]
   \[
   \sqrt[3]{10000} = 21.54
   \]

9. Since \( n \) must be an integer, the only possible value for \( n \) within this range is \( n = 21 \).

10. Verify:
    \[
    21^3 = 9261
    \]
    which is indeed between 9000 and 10000.

The final answer is \( \boxed{21} \).