7. Given a positive integer $n$ does not exceed 2000, and can be expressed as the sum of no less than 60 consecutive positive integers. Then, the number of such $n$ is $\qquad$ .
Since $1+2+\cdots+62=63 \times 31=1953$, we have $n=\sum_{k=1}^{60} k, \sum_{k=1}^{61} k, \sum_{k=1}^{62} k, \sum_{k=2}^{61} k, \sum_{k=2}^{62} k$,
$$
\sum_{k=3}^{62} k \text { for a total of } 6 \text { terms. }
$$