10. (20 points) Let $x \in (0,1), \frac{1}{x} \notin \mathbf{Z}$,
$$
a_{n}=\frac{n x}{(1-x)(1-2 x) \cdots(1-n x)},
$$

where $n=1,2, \cdots$.
We call $x$ a "good number" if and only if $x$ makes the sequence $\left\{a_{n}\right\}$ defined above satisfy
$$
a_{1}+a_{2}+\cdots+a_{10}>-1 \text { and } a_{1} a_{2} \cdots a_{10}>0 \text {. }
$$

Find the sum of the lengths of all intervals on the number line corresponding to all good numbers.
10. From the given conditions, we have
$$
\begin{array}{l}
\sum_{n=1}^{10} a_{n}=\sum_{n=1}^{10}\left(\frac{1}{(1-x) \cdots(1-n x)}-\right. \\
\left.\frac{1}{(1-x) \cdots(1-(n-1) x)}\right) \\
=\frac{1}{(1-x)(1-2 x) \cdots(1-10 x)}-1 \text {. } \\
\end{array}
$$

Thus, \( x \) is a good number if and only if, for \( x \in (0,1) \) and \( \frac{1}{x} \notin \mathbf{Z} \), we have
$$
\begin{array}{l}
(1-x)(1-2 x) \cdots(1-10 x)>0, \\
\prod_{n=1}^{10} a_{n}=\frac{10! \cdot x^{10}}{(1-x)^{10}(1-2 x)^{9} \cdots(1-10 x)} \\
>0 .
\end{array}
$$

This is equivalent to
$$
\begin{array}{l}
x \in (0,1), \frac{1}{x} \notin \mathbf{Z}, \\
(x-1)\left(x-\frac{1}{2}\right) \cdots\left(x-\frac{1}{10}\right)>0, \\
\left(x-\frac{1}{2}\right)\left(x-\frac{1}{4}\right) \cdots\left(x-\frac{1}{10}\right)<0 .
\end{array}
$$

Solving this, we get \( x \in \left(0, \frac{1}{10}\right) \cup \left(\frac{1}{7}, \frac{1}{6}\right) \cup \left(\frac{1}{3}, \frac{1}{2}\right) \),
$$
x \neq \frac{1}{11}, \frac{1}{12}, \cdots \text {. }
$$

Notice that removing countably many points from the interval \(\left(0, \frac{1}{10}\right)\) does not affect the total length of the intervals. Therefore, the sum of the lengths of the intervals is
$$
\frac{1}{10}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{19}{210} .
$$