8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.

Then $m=$ $\qquad$ $(\bmod 2011)$.
8. 0 .

Notice
$$
\begin{array}{l} 
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\
= 2011 \times \frac{q}{2010!},
\end{array}
$$

where $q$ is a positive integer.
Thus, $2011 n q=m \times 2010!$.
This shows that 2011 divides $m \times 2010!$, but 2011 is a prime number and does not divide 2010!, so 2011 must divide $m$, giving $m \equiv 0(\bmod 2011)$.