5. (1999 National High School Competition Question) 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 $\qquad$ .
5.6. The sum of the first term $a$ of $k$ consecutive positive integers is $S_{k}=k a+\frac{1}{2} k(k-1) \geqslant \frac{k(k+1)}{2}$.

From $S_{k} \leqslant 2000$ we get $60 \leqslant k \leqslant 62 . k=60$ when $S_{k}=60 a+30 \cdot 59 \leqslant 2000$, we get $a \leqslant 3$, so $S_{k}=1830,1890$, 1950; when $k=61$, $S_{k}=61 a+30 \cdot 61 \leqslant 2000$, we get $a \leqslant 2$, so $S_{k}=1891,1952$; when $k=62$, $S_{k}=62 a+31 \cdot$ $61 \leqslant 2000$, we get $a \leqslant 1$, so $S_{k}=1953$. Thus, there are 6 values for $n$ in the problem.