8. Let the positive divisors of $2014^{2}$ be $d_{1}, d_{2}, \cdots$, $d_{k}$. Then
$$
\frac{1}{d_{1}+2014}+\frac{1}{d_{2}+2014}+\cdots+\frac{1}{d_{k}+2014}
$$
$=$ . $\qquad$
8. $\frac{27}{4028}$.

Notice that, $2014^{2}=2^{2} \times 19^{2} \times 53^{2}$.
Thus, $2014^{2}$ has $(2+1)^{3}=27$ positive divisors, i.e., $k=27$.
Let $1=d_{1}<d_{2}<\cdots<d_{27}=2014^{2}$.
Therefore, $d_{i} d_{28-i}=2014^{2}(i=1,2, \cdots, 27)$.
Then $\frac{1}{d_{i}+2014}+\frac{1}{d_{28-i}+2014}$
$$
\begin{array}{l}
=\frac{d_{i}+d_{28-i}+2 \times 2014}{d_{i} d_{28-i}+2014\left(d_{i}+d_{28-i}\right)+2014^{2}} \\
=\frac{1}{2014} .
\end{array}
$$
$$
\begin{array}{c}
\text { Hence } \frac{1}{d_{1}+2014}+\frac{1}{d_{2}+2014}+\cdots+\frac{1}{d_{27}+2014} \\
=\frac{1}{2}\left[\left(\frac{1}{d_{1}+2014}+\frac{1}{d_{27}+2014}\right)+\right. \\
\left(\frac{1}{d_{2}+2014}+\frac{1}{d_{26}+2014}\right)+\cdots+ \\
\left.\left(\frac{1}{d_{27}+2014}+\frac{1}{d_{1}+2014}\right)\right] \\
=\frac{1}{2} \times 27 \times \frac{1}{2014}=\frac{27}{4028} .
\end{array}
$$