Question 213, Solution: Let's assume the first three decimal places of the number are $\overline{251}$, which is equivalent to
$$
251 \leq \frac{1000 \mathrm{~m}}{\mathrm{n}}\frac{1000-252 \mathrm{r}}{8} \geq \frac{1000-252 \times 3}{8}>30 \\
\Rightarrow \mathrm{k} \geq 31
$$

So $\mathrm{k} \geq 31$, and the equality can only hold when $r=3$. Therefore, $n=4 k+r \geq 4 \times 31+3=127$.
It is also easy to verify that the first three decimal places of $\frac{32}{127}$ are $\overline{251}$, so the smallest value of $\mathrm{n}$ that satisfies the condition is 127.
Question 213: If $m, n \in Z^{+}, (m, n)=1, m<n$, and the decimal representation of $\frac{\mathrm{m}}{\mathrm{n}}$ contains $\overline{251}$, find the minimum value of $\mathrm{n}$.