8. Given that $a, b, c, d$ are positive numbers, and
$$
a+20 b=c+20 d=2 \text {. }
$$

Then the minimum value of $\frac{1}{a}+\frac{1}{b c d}$ is
8. $\frac{441}{2}$.

From the condition, we have
$$
\begin{aligned}
0 < c d & =\frac{1}{20} \cdot c \cdot 20 d \leqslant \frac{1}{20}\left(\frac{c+20 d}{2}\right)^{2}=\frac{1}{20}. \\
& \text{Then } \frac{1}{a}+\frac{1}{b c d} \geqslant \frac{1}{a}+\frac{20}{b} \\
& =\frac{1}{2}\left(\frac{1}{a}+\frac{20}{b}\right)(a+20 b) \\
& =\frac{1}{2}\left(401+\frac{20 b}{a}+\frac{20 a}{b}\right) \\
& \geqslant \frac{1}{2}\left(401+2 \sqrt{\frac{20 b}{a} \cdot \frac{20 a}{b}}\right) \\
& =\frac{441}{2},
\end{aligned}
$$

Equality holds if and only if $c=20 d$ and $\frac{20 b}{a}=\frac{20 a}{b}$, i.e.,
$$
a=b=\frac{2}{21}, c=1, d=\frac{1}{20}
$$

Thus, the minimum value of $\frac{1}{a}+\frac{1}{b c d}$ is $\frac{441}{2}$.