6. Given that $n$ is a positive integer, $a_{1}, a_{2}, \cdots, a_{n}$ are real numbers not all zero, define
$$
x=\frac{\left|a_{1}+a_{2}\right|+\left|a_{2}+a_{3}\right|+\cdots+\left|a_{n}+a_{1}\right|}{\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|} \text {. }
$$

Then the maximum value of $x$ is , and the minimum value of $x$ is
$6.2, \frac{1-(-1)^{n}}{n}$.
First, find the maximum value.
Notice,
$$
\begin{array}{l}
\left|x_{1}+x_{2}\right| \leqslant\left|x_{1}\right|+\left|x_{2}\right|, \\
\left|x_{2}+x_{3}\right| \leqslant\left|x_{2}\right|+\left|x_{3}\right|, \\
\cdots \cdots \cdots \\
\left|x_{n}+x_{1}\right| \leqslant\left|x_{n}\right|+\left|x_{1}\right| .
\end{array}
$$
$$
\begin{array}{l}
\text { Then }\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right| \\
\leqslant 2\left(\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|\right) . \\
\text { Hence } \frac{\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right|}{\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|} \leqslant 2 .
\end{array}
$$

When $x_{1}=x_{2}=\cdots=x_{n}$, the equality holds.
Next, find the minimum value.
Discuss the parity of $n$.
If $n$ is odd, then
$$
\begin{array}{l}
\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right| \\
\geqslant \mid\left(x_{1}+x_{2}\right)-\left(x_{2}+x_{3}\right)+\left(x_{3}+x_{4}\right)- \\
\cdots+(-1)^{n-1}\left(x_{n}+x_{1}\right) \mid \\
= 2\left|x_{1}\right| .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right| \\
\geqslant 2\left|x_{2}\right|, \\
\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right| \\
\geqslant 2\left|x_{3}\right|, \\
\cdots \cdots \\
\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right| \\
\geqslant 2\left|x_{n}\right| .
\end{array}
$$

Adding the above inequalities, we get
$$
\begin{array}{l}
n\left(\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right|\right) \\
\geqslant 2\left(\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|\right) . \\
\text { Hence } \frac{\left|x_{1}+x_{2}\right|+\left|x_{2}+x_{3}\right|+\cdots+\left|x_{n}+x_{1}\right|}{\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|} \\
\geqslant \frac{2}{n} . \\
\text { When } x_{1}=-x_{2}=\cdots=(-1)^{n-1} x_{n} \text {, the equality holds. }
\end{array}
$$

Similarly, if $n$ is even, the minimum value is 0.