13. (15 points) Let $[x]$ denote the greatest integer not exceeding $x$. Suppose a natural number $n$ satisfies
$$
\left[\frac{1}{15}\right]+\left[\frac{2}{15}\right]+\left[\frac{3}{15}\right]+\cdots+\left[\frac{\mathrm{n}-1}{15}\right]+\left[\frac{\mathrm{n}}{15}\right]>2000 \text {, }
$$

then what is the minimum value of $n$?
【Analysis】When $n$ is a multiple of 15, the integer quotients obtained are $1, 2, 3, \cdots k$, so the first 14 fractions rounded down are all 0, and from $\frac{15}{15}=1$ onwards, let the rounded down values of every 15 fractions be $1, 2, 3, \cdots k$, so we can get: 14 $\times 0+15 \times 1+15 \times 2+15 \times 3+\cdots+15 \times k=15 \times(1+2+3+\cdots+k)>2000$, which means $(k+1) \times k>\frac{800}{3}$, then we can discuss the range of $k$.

【Solution】When $n$ is a multiple of 15, the first 14 fractions rounded down are all 0, and from $\frac{15}{15}=1$ onwards, let the rounded down values of every 15 fractions be $1, 2, 3, \cdots k$, so we can get:
$$
14 \times 0+15 \times 1+15 \times 2+15 \times 3+\cdots+15 \times k=15 \times(1+2+3+\cdots+k)>2000,
$$

which means $(k+1) \times k>\frac{800}{3}$, then $(k+1) \times k \geqslant 267$,
First, find the nearest integer,
then, $15 \times(15+1)=240$
$$
(16+1) \times 16=272 \text {, }
$$

When $k=16$, i.e., $n=15 \times(16-1)+1+14=240$
then, the sum of the rounded down values of the numbers before $\frac{240}{15}$ is:
i.e., $15 \times(1+2+3+\cdots+15)=1800$,
$1800<2000$,
still need $2000-1800-\frac{240}{15}=184$,
So, we need to take another $184 \div 16 \approx 12$,
So, the minimum value of $n$ is: $240+12=252$.
Answer: The minimum value of $n$ is 252.