A34 (36-4, Poland) Let the sequence of positive real numbers $x_{0}, x_{1}, \cdots, x_{1995}$ satisfy the following two conditions:
(1) $x_{0}=x_{1995}$;
(2) $x_{i, 1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}, i=1,2, \cdots, 1995$.
Find the maximum value of $x_{0}$ for all sequences satisfying the above conditions.
Solve: Rewrite the recurrence relation as
$$
\left(x_{i-1}-2 x_{i}\right)+\frac{1}{x_{i} x_{i-1}}\left(2 x_{i}-x_{i-1}\right)=0 \text {. }
$$

which is
$$
\left(2 x_{i}-x_{i-1}\right)\left(\frac{1}{x_{i} x_{i-1}}-1\right)=0 \text {. }
$$

From this, we know
$$
x_{i}=\frac{x_{i-1}}{2} \text { or } x_{i}=\frac{1}{x_{1} 1} .
$$

Assume starting from $x_{0}$, it undergoes $k-t$ transformations of the first type and $t$ transformations of the second type to get $x_{k}$, it is easy to see
$$
x_{k}=2 x_{0}^{(-1)^{\prime}} .
$$

where $s \equiv k-t \quad(\bmod 2)$.
Now consider the case $k=1995$, if $t$ is even, then
$$
x_{0}=x_{1995}=2^{*} x_{0} .
$$

Since $x_{0}>0$, we get $s=0$, i.e.,
$$
0 \equiv 1995-t \quad(\bmod 2),
$$

which contradicts $t$ being even. Therefore, $t$ must be odd. Thus,
$$
x_{0}=x_{1995}=2^{*} x_{0}^{-1} \text {, }
$$

which gives
$$
x_{0}=2^{\frac{1}{2}} .
$$

In this case, $s \equiv 1995-t \equiv 0(\bmod 2)$, so $s$ is even. When the two types of transformations are alternated, there might be a cancellation effect, hence $s \leqslant|s| \leqslant 1995$.
Thus, $s \leqslant 1994, x_{0} \leqslant 2^{997}$.
When $x_{0}=2^{997}$, the recurrence can be realized. That is,
$$
\begin{array}{c}
x_{i}=2^{997-i}, i=0,1,2, \cdots, 1994 ; \\
x_{1995}=\frac{1}{x_{1994}}=2^{997} .
\end{array}
$$

In summary, the maximum value of $x_{0}$ is $2^{997}$.