6. Given an integer $n \geqslant 2$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\left|x_{i}\right| \leqslant 1(i=1,2, \cdots, n)$.

Find the maximum value of $\min \left|x_{i}-x_{i+1}\right|$.
6. Let $A=\min _{1 \leqslant i \leqslant n-1}\left|x_{i}-x_{i+1}\right|$.
(i) When $n$ is even, from condition (2) we know that for $1 \leqslant i \leqslant n-1$, we have
$$
\left|x_{i}-x_{i+1}\right| \leqslant\left|x_{i}\right|+\left|x_{i+1}\right| \leqslant 2 \text {. }
$$

Thus, $A \leqslant 2$.
Taking $x_{i}=(-1)^{i}(i=1,2, \cdots, n)$, satisfies conditions (1) and (2), and $A=2$.
Therefore, the maximum value of $A$ is 2.
(ii) When $n$ is odd, let $n=2 k+1$.
If there exists $i$ such that $x_{i} \leqslant x_{i+1} \leqslant x_{i+2}$ or $x_{i} \geqslant x_{i+1} \geqslant x_{i+2}$, then by $A \leqslant\left|x_{i}-x_{i+1}\right|, A \leqslant\left|x_{i+1}-x_{i+2}\right|$, we have
$$
2 A \leqslant\left|x_{i+2}-x_{i}\right| \leqslant 2 \text {. }
$$

Thus, $A \leqslant 1x_{2 i}, x_{2 i}<x_{2 i+1}$, where $i=1,2, \cdots, k$, then,
$$
\begin{aligned}
& (2 k+2) A \\
& \leqslant \sum_{i=1}^{k}\left(\left|x_{2 i-1}-x_{2 i}\right|+\left|x_{2 i}-x_{2 i+1}\right|\right)+ \\
= & \sum_{i=1}^{k}\left(\mid x_{2 i-1}-x_{2 i}+x_{2 i+1}-x_{2 i}\right)+ \\
& \left(x_{1}-x_{2}+x_{2 k+1}-x_{2 k}\right) \\
= & 2 \sum_{i=1}^{k+1} x_{2 i-1}-2 \sum_{i=1}^{k} x_{2 i}-x_{2}-x_{2 k} \\
= & -4 \sum_{i=1}^{k} x_{2 i}-x_{2}-x_{2 k} \\
\leqslant & 4 k+2 \\
\Rightarrow & A \leqslant \frac{4 k+2}{2 k+2}=\frac{2 n}{n+1} . \\
\text { Taking } & x_{i}=\left\{\begin{array}{c}
\frac{k}{k+1}, \quad i=1,3, \cdots, 2 k+1 ; \\
-1, \quad i=2,4, \cdots, 2 k .
\end{array}\right.
\end{aligned}
$$

It is easy to verify that this sequence satisfies conditions (1) and (2), and
$$
A=\frac{2 k+1}{k+1}=\frac{2 n}{n+1} \text {. }
$$

Therefore, the maximum value of $A$ is $\frac{2 n}{n+1}$.