6. Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=\sqrt{3}, x_{n+1}=\frac{x_{n}+1}{1-x_{n}} \text {. }
$$

Then $x_{1001}-x_{401}=$
6.0 .

From the structural characteristics, we can make the substitution $x_{n}=\tan a_{n}$. Then $x_{n+1}=\frac{\tan a_{n}+\tan \frac{\pi}{4}}{1-\tan a_{n} \cdot \tan \frac{\pi}{4}}=\tan \left(a_{n}+\frac{\pi}{4}\right)$
$$
\begin{array}{c}
\Rightarrow x_{n+2}=\tan \left[\left(a_{n}+\frac{\pi}{4}\right)+\frac{\pi}{4}\right] \\
=\tan \left(a_{n}+2 \times \frac{\pi}{4}\right) .
\end{array}
$$

Similarly, $x_{n+3}=\tan \left(a_{n}+3 \times \frac{\pi}{4}\right)$,
$$
\begin{array}{l}
x_{n+4}=\tan \left(a_{n}+4 \times \frac{\pi}{4}\right) \\
=\tan \left(a_{n}+\pi\right)=\tan a_{n}=x_{n} .
\end{array}
$$

Thus, the sequence $\left\{x_{n}\right\}$ has a period of 4.
Since $1001=150 \times 4+401$, we have,
$$
x_{1001}-x_{401}=0 \text {. }
$$