Let $n$ be a positive integer such that $1\le n\le 1000$. Let $M_n$ be the number of perfect squares in the set $X_n=\{4n+1,4n+2,\dots ,4n+1000\}$. Let $a=\max \{M_n:1\le n\le 1000\}$ and $b=\min \{M_n:1\le n\le 1000\}$. Find $a-b$.

Step 1: The spacing between consecutive perfect squares increases as the numbers grow. Therefore, the set $X_n$ will contain more perfect squares when $n$ is small (squares are closer together) and fewer perfect squares when $n$ is large (squares are further apart).

Step 2: Thus, the maximum count of perfect squares occurs at the minimum value of $n$, so $a=M_1$, and the minimum count occurs at the maximum value of $n$, so $b=M_{1000}$.

Step 3: For $n=1$, the set is $X_1=\{5,6,\dots ,1004\}$. The perfect squares in this range are $3^2=9,4^2=16,\dots ,31^2=961$. The number of squares is $a=31-3+1=29$.

Step 4: For $n=1000$, the set is $X_{1000}=\{4001,4002,\dots ,5000\}$. The perfect squares in this range are $64^2=4096,65^2=4225,\dots ,70^2=4900$. The number of squares is $b=70-64+1=7$.

Step 5: Finally, we compute $a-b$: