12. (10 points) Let $[x]$ denote the greatest integer not exceeding $x$. Suppose a natural number $n$ satisfies $\left[\frac{1}{15}\right]+\left[\frac{2}{15}\right]+\left[\frac{3}{15}\right]+\cdots$ $+\left[\frac{\mathrm{n}-1}{15}\right]+\left[\frac{\mathrm{n}}{15}\right]>2011$, then what is the minimum value of $n$?
【Analysis】Observation: $\left[\frac{1}{15}\right]=0,\left[\frac{2}{15}\right]=0, \cdots,\left[\frac{14}{15}\right]=0$, the sum of the first 14 numbers is 0 $\left[\frac{15}{15}\right]=1,\left[\frac{16}{15}\right]=\left[1 \frac{1}{15}\right]=1, \cdots,\left[\frac{29}{15}\right]=\left[1 \frac{14}{15}\right]=1$, these 15 numbers are all 1, the sum is $1 \times 15=$ 15,
$\left[\frac{30}{15}\right]=2,\left[\frac{31}{15}\right]=\left[2 \frac{1}{15}\right]=2, \cdots,\left[\frac{44}{15}\right]=\left[2 \frac{14}{15}\right]=2$, these 15 numbers are all 2, the sum is $2 \times 15=$ 30,
$\cdots$
Observation can show that the pattern is to increase by 1 every 15 numbers,
$$
\begin{array}{l}
(1+2+3+\cdots+15) \times 15=1800, \\
(1+2+3+\cdots+15+16) \times 15=2040,2040>2011,
\end{array}
$$

Therefore, the integer part adds up to 15, just reaching 1800, continuing to the integer part of 16, $2011-1800=211,211 \div 16=13.1875$, so at least 14 need to be taken, i.e., the minimum is to take up to $16 \frac{13}{15}$, to ensure it is greater than 2011,

Then the minimum value of $n$ is: $16 \times 15+13=253$.
【Solution】Solution:
$$
(1+2+3+\cdots+15) \times 15=1800,
$$
$(1+2+3+\cdots+15+16) \times 15=2040,2040>2011$, then the integer part is 16,
$2011-1800=211$,
$211 \div 16=13.1875$
i.e., at least up to $16 \frac{13}{15}$, to ensure it is greater than 2011,
Then the minimum value of $n$ is: $16 \times 15+13=253$.
Answer: The minimum value of the natural number $n$ is 253.