Example 13 Let the sequence of positive real numbers $x_{0}, x_{1}, x_{2}, \cdots, x_{1995}$ satisfy the conditions:
(1) $x_{0}=x_{1995}$;
(2) For $i=1,2, \cdots, 1995$, $x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}$, find the maximum value of $x_{0}$.
Solution: When $x_{i}$ are all equal to 1, the condition is satisfied, so if there exists the required maximum value $x_{0}$, it must be greater than or equal to 1. When $x_{i-1}(i \geqslant 1)$ is fixed, by condition (2), $x_{i}$ must satisfy the quadratic equation
$$
2 x^{2}-\left(x_{i-1}+\frac{2}{x_{i-1}}\right) x+1=\left(2 x-x_{i-1}\right)\left(x-\frac{1}{x_{i-1}}\right)=0,
$$

Therefore, it must be
$$
x_{i}=\frac{1}{2} x_{i-1} \text { or } x_{i}=\frac{1}{x_{i-1}} \text {. }
$$

It is evident that these two values both satisfy condition (2). Therefore, starting from the initial value $x_{0}$, the numbers in the sequence must be of the following form, i.e.,
$$
2^{k} x_{0}, 2^{k} x_{0}^{-1}, k \in \mathbf{Z} .
$$

Since $x_{0} \geqslant 1$, to satisfy (1), the possible values of $x_{1995}$ are
$$
x_{0}\left(2^{k} x_{0} \text { with } k=0\right) \text { or } 2^{k} x_{0}^{-1}(k \in \mathbf{Z}) \text {. }
$$

We will now prove that $x_{1995}$ can only take the form $2^{k} x_{0}^{-1}$, and the maximum possible value of $k$ is 1994. Therefore, the maximum value sought in this problem is $x_{0}=2^{997}$.

We first analyze the possible values of the sequence that satisfy condition (2). We use a diagram to represent this. After $x_{0}$ is determined, the possible values of $x_{1}$ are shown by the arrow in Figure 4-5.
If $x_{1}=x_{0}^{-1}$, then the possible values of $x_{2}$ are shown in Figure 4-6.

If $x_{1}=2^{-1} x_{0}$, then the possible values of $x_{2}$ are shown in Figure 4-7.
Combining these three figures, we get Figure 4-8.
Therefore, the possible values of the sequence $x_{0}, x_{1}, x_{2}$ can be obtained by starting from $x_{0}$ in Figure 4-8 and sequentially taking the next value along the direction indicated by the arrow, resulting in four possible sequences. By this logic, the possible value sequences of $x_{0}, x_{1}, \cdots, x_{n}$ can be represented by such a diagram: starting from $x_{0}$ in Figure 4-9 and sequentially taking the next value along the direction indicated by the arrow until $x_{n}$. Clearly, there are $2^{n}$ possible paths. For example,
$$
\begin{array}{c}
x_{0}, 2^{-1} x_{0}, 2 x_{0}^{-1}, x_{0}^{-1}, x_{0}, 2^{-1} x_{0}, 2^{-2} x_{0}, 2^{2} x_{0} \\
x_{0}, x_{0}^{-1}, 2^{-1} x_{0}^{-1}, 2^{-2} x_{0}^{-1}, 2^{2} x_{0}, 2 x_{0}, x_{0}
\end{array}
$$

respectively give two sequences of values for $n=7$ and $n=6$.
From the structure of Figure 4-8, it is easy to see that if a certain value path of the sequence $x_{0}, x_{1}, \cdots, x_{n}$ indeed returns to $x_{0}$ in the diagram, then the value path of this sequence must be represented by a closed loop (which can be repeated). Noting that $x_{n}=x_{0}$, the number of values taken in the first row must be one more than the number of values taken in the second row. Therefore, $n$ must be even. Now $n=1995$, so this scenario is impossible. Therefore, it can only be
$$
x_{0}=x_{n}=2^{k} x_{0}^{-1}
$$

Clearly, the larger $k$ is, the larger $x_{0}$ is. From the diagram, the maximum value of $k$ is 1994. This sequence can only be
$$
\left\{\begin{array}{l}
x_{i}=2^{-i} x_{0}, 0 \leqslant i \leqslant 1994 \\
x_{1995}=2^{1994} x_{0}^{-1}
\end{array}\right.
$$

i.e., $x_{0}=2^{997}$. Proof complete.