II. Fill-in-the-blank Questions (Full marks 54 points, each question 9 points)  
1. Given a positive integer $n$ does not exceed 2000, and can be expressed as the sum of at least 60 consecutive positive integers, then the number of such $n$ is $\qquad$.
6
1. 【Analysis and Solution】The sum of $k$ consecutive positive integers with the first term $a$ is $S_{k}=k a+\frac{k(k+1)}{2} \geqslant \frac{k(k+1)}{2}$. From $S_{k} \leqslant 2000$, we can get $60 \leqslant k \leqslant 62$.
When $k=60$, $S_{k}=60 a+30 \times 59$, from $S_{k} \leqslant 2000$, we can get $a \leqslant 3$, so $S_{k}=1830,1890,1950$; When $k=61$, $S_{k}=61 a+30 \times 61$, from $S_{k} \leqslant 2000$, we can get $a \leqslant 2$, so $S_{k}=1891,1952$; When $k=62$, $S_{k}=62 a+31 \times 61$, from $S_{k} \leqslant 2000$, we can get $a \leqslant 1$, so $S_{k}=1953$.
$\therefore$ There are 6 values for $n$ in the problem.