Let $n$ be a positive integer such that $1 \leq n \leq 1000$. Let $M_n$ be the number of integers in the set $X_n=\{\sqrt{4 n+1}, \sqrt{4 n+2}, \ldots, \sqrt{4 n+1000}\}$. Let
$$
a=\max \left\{M_n: 1 \leq n \leq 1000\right\} \text {, and } b=\min \left\{M_n: 1 \leq n \leq 1000\right\} \text {. }
$$
Find $a-b$.
1. We need to determine the number of integers in the set \( X_n = \{\sqrt{4n+1}, \sqrt{4n+2}, \ldots, \sqrt{4n+1000}\} \) that are perfect squares. This means we need to find the number of perfect squares in the interval \([4n+1, 4n+1000]\).

2. Let \( k^2 \) be a perfect square in the interval \([4n+1, 4n+1000]\). Then:
   \[
   4n+1 \leq k^2 \leq 4n+1000
   \]
   This implies:
   \[
   \sqrt{4n+1} \leq k \leq \sqrt{4n+1000}
   \]

3. To find the number of integers \( k \) that satisfy this inequality, we need to determine the integer values of \( k \) between \(\sqrt{4n+1}\) and \(\sqrt{4n+1000}\).

4. Let \( a = \lfloor \sqrt{4n+1} \rfloor \) and \( b = \lfloor \sqrt{4n+1000} \rfloor \). The number of integers \( k \) in this range is:
   \[
   M_n = b - a + 1
   \]

5. We need to find the maximum and minimum values of \( M_n \) for \( 1 \leq n \leq 1000 \).

6. For \( n = 1 \):
   \[
   \sqrt{4 \cdot 1 + 1} = \sqrt{5} \approx 2.236 \quad \text{and} \quad \sqrt{4 \cdot 1 + 1000} = \sqrt{1004} \approx 31.654
   \]
   Thus, \( a = 2 \) and \( b = 31 \), so:
   \[
   M_1 = 31 - 2 + 1 = 30
   \]

7. For \( n = 1000 \):
   \[
   \sqrt{4 \cdot 1000 + 1} = \sqrt{4001} \approx 63.245 \quad \text{and} \quad \sqrt{4 \cdot 1000 + 1000} = \sqrt{5000} \approx 70.711
   \]
   Thus, \( a = 63 \) and \( b = 70 \), so:
   \[
   M_{1000} = 70 - 63 + 1 = 8
   \]

8. Therefore, the maximum value of \( M_n \) is 30 and the minimum value of \( M_n \) is 8. Hence:
   \[
   a = 30 \quad \text{and} \quad b = 8
   \]

9. The difference \( a - b \) is:
   \[
   a - b = 30 - 8 = 22
   \]

The final answer is \(\boxed{22}\).