Given a positive integer $n$, solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+1=\frac{1}{x_{2}} \\
\ldots \ldots \\
x_{n-1}+1=\frac{1}{x_{n}} \\
x_{n}+1=\frac{1}{x_{1}}
\end{array}\right.
$$
$$
x_{1}=x_{2}=\cdots=x_{n}=\frac{-1-\sqrt{5}}{2} \text { or } \frac{-1+\sqrt{5}}{2}
$$

are solutions to the system of equations.
Next, we prove: $x_{1}=x_{2}=\cdots=x_{n}$.
Subtracting the first two equations, we get
$$
x_{1}-x_{2}=\frac{1}{x_{2}}-\frac{1}{x_{3}}=\frac{x_{3}-x_{2}}{x_{2} x_{3}}.
$$

Similarly, $x_{i}-x_{i+1}=\frac{x_{i+2}-x_{i+1}}{x_{i+1} x_{i+2}}$.
Multiplying all these equations, we get
$$
\begin{array}{l}
\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right) \cdots\left(x_{n}-x_{1}\right) \\
=\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right) \cdots \\
\quad\left(x_{n}-x_{1}\right)(-1)^{n} \frac{1}{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}},
\end{array}
$$

which simplifies to $\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right) \cdots\left(x_{n}-x_{1}\right)\left(1 \pm \frac{1}{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}}\right) = 0$.
Clearly, $1+\frac{1}{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}} \neq 0$.
We now prove: $1-\frac{1}{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}} \neq 0$.
Assume, for contradiction, that $x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}=1$.
Multiplying the given $n$ equations, we get
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right) \cdots\left(x_{n}+1\right) \\
=\frac{1}{x_{1} x_{2} \cdots x_{n}}=x_{1} x_{2} \cdots x_{n}.
\end{array}
$$

If all $x_{i}>0$, then $x_{i}+1>x_{i} \geqslant 0 (i=1,2, \cdots, n)$. Multiplying these $n$ inequalities contradicts equation (1).

If at least one of $x_{1}, x_{2}, \cdots, x_{n}$ is negative, assume without loss of generality that $x_{1}-x_{i}-1>0$.
Multiplying these $n$ inequalities contradicts equation (1).
Thus, $1 \pm \frac{1}{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}} \neq 0$.
Therefore, $\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right) \cdots\left(x_{n}-x_{1}\right)=0$.
Assume without loss of generality that $x_{1}=x_{2}$. From the first two equations, we get $\frac{1}{x_{2}}=\frac{1}{x_{3}}$.
Thus, $x_{2}=x_{3}$.
Continuing this, we have $x_{1}=x_{2}=\cdots=x_{n}$.
From the system of equations, we get $x_{1}+1=\frac{1}{x_{1}}$, i.e., $x_{1}^{2}+x_{1}=1$.
Solving, we get $x_{1}=\frac{-1 \pm \sqrt{5}}{2}$.

Thus, the solutions to the system of equations are
$$
\begin{aligned}
( & \left.x_{1}, x_{2}, \cdots, x_{n}\right) \\
= & \left(\frac{-1-\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \cdots, \frac{-1-\sqrt{5}}{2}\right), \\
& \left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}, \cdots, \frac{-1+\sqrt{5}}{2}\right).
\end{aligned}
$$