6. $n \geqslant 3$, the equation $\frac{x_{1} x_{2} \cdots x_{n-1}}{x_{n}}+\frac{x_{1} x_{2} \cdots x_{n-2} x_{n}}{x_{n-1}}$ $+\cdots+\frac{x_{2} x_{3} \cdots x_{n}}{x_{1}}=n$ has the following number of ordered integer solutions:
(A) $n$ groups
(B) $2^{n-1}$ groups
(C) $2^{n}$ groups
(D) $2^{n+1}$ groups
6. (B).
$x_{i} \neq 0(i=1,2, \cdots, n)$, from the original equation we have
$$
\begin{array}{l}
x_{1}^{2} x_{2}^{2} \cdots x_{n-1}^{2}+x_{1}^{2} x_{2}^{2} \cdots x_{n-2}^{2} x_{n}^{2} \cdots \cdots \\
+x_{2}^{2} x_{3}^{2} \cdots x_{n}^{2} \\
=n x_{1} x_{2} \cdots x_{n} .
\end{array}
$$

Then, $x_{1} x_{2} \cdots x_{n}>0$.
The left side $\geqslant_{n}\left(x_{1} x_{2} \cdots x_{n}\right)^{\frac{2(n-1)}{n}}$.
Thus, we have
$$
\begin{array}{l}
\left(x_{1} x_{2} \cdots x_{n}\right)^{n} \geqslant\left(x_{1} x_{2} \cdots x_{n}\right)^{2(n-1)} . \\
\left(x_{1} x_{2} \cdots x_{n}\right)^{n-2} \leqslant 1 .
\end{array}
$$

Since $n \geqslant 3$, then $x_{1} x_{2} \cdots x_{n}=1$. The equation has solutions $(1,1, \cdots, 1),(1,1, \cdots, 1,-1,-1)$, etc., with a total of $C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+C_{n}^{\overline{2\left(\frac{n}{2}\right]}}$ solutions. When $n=2 k$, the sum is $2^{2 k-1}$; when $n=2 k+1$, the sum is $2^{2 k}$. Therefore, there are $2^{n-1}$ solutions.