8. Given $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$ has a segment of digits $\overline{2012}$ in its decimal part, where $n$ is the smallest number satisfying the condition. Then $\left[\frac{m}{\sqrt{n}}\right]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
8. 2 .

Let $\frac{m}{n}=\overline{A . B 2012 C}$, where $A, B, C$ are digit strings, and the lengths of $A, B$ are $k, l (k, l \in \mathbf{N})$. Then
$$
\frac{10^{6}(m-n A)-n B}{n}=\overline{0.2012 C} \triangleq \frac{a}{b},
$$

where $(a, b)=1, 1 \leqslant b \leqslant n$.
Assume $\frac{m}{n}=\overline{0.2012 C}$.
$$
\begin{array}{l}
\text { Hence } \frac{2012}{10000} \leqslant \frac{m}{n}<\frac{2013}{10000} \\
\Rightarrow \frac{10000}{2013}<\frac{n}{m} \leqslant \frac{10000}{2012} .
\end{array}
$$

Let $n=5 m-r$. From
$$
\begin{array}{l}
\text { Equation (1) } \Rightarrow \frac{60}{2.012} \leqslant \frac{r}{m}<\frac{65}{2013} \\
\Rightarrow \frac{2013}{13}<\frac{5 m}{r}<\frac{503}{3} \\
\Rightarrow \frac{2000}{13} r<5 m-r \leqslant \frac{500}{3} r .
\end{array}
$$

Take $r=1, m=31$, then $n=5 m-r=154$. This is the minimum value.
Thus $\left[\frac{m}{\sqrt{n}}\right]=\left[\sqrt{\frac{961}{154}}\right]=2$.