24. S2 (POL) $)^{1 \mathrm{MO}}$ The positive real numbers $x_{0}, x_{1}, \ldots, x_{1996}$ satisfy $x_{0}=$ $x_{1995}$ and
$$
x_{i-1}+\frac{2}{x_{i-1}}=2 x_{i}+\frac{1}{x_{i}}
$$
for $i=1,2, \ldots, 1995$. Find the maximum value that $x_{0}$ can have.
24. The given condition is equivalent to $\left(2 x_{i}-x_{i-1}\right)\left(x_{1} x_{i-1}-1\right)=0$, so either $x_{i}=\frac{1}{2} x_{i-1}$ or $x_{i}=\frac{1}{x_{i-1}}$. We shall show by induction on $n$ that for any $n \geq 0, x_{n}=2^{k_{n}} x_{0}^{e_{0}}$ for some integer $k_{n}$, where $\left|k_{n}\right| \leq n$ and $e_{n}=(-1)^{\bar{n}-k_{n}}$. Indeed, this is true for $n=0$. If it holds for some $n$, then $x_{n+1}=\frac{1}{2} x_{n}=2^{k_{n}-1} x_{0}^{2}$ (hence $k_{n+1}=k_{n}-1$ and $e_{n+1}=e_{n}$ ) or $x_{n+1}=\frac{1}{x_{n}}=2^{-k_{n}} x_{0}^{-\varepsilon_{n}}$ (hence $k_{n+1}=-k_{n}$ and $e_{n+1}=-e_{n}$ ).
Thus $x_{0} x_{n}=x_{1995}=2^{k_{19}} x_{0}^{\varepsilon 1905}$. Note that $e_{1995}=1$ is impossible, since in that case $\hbar_{1995}$ would be odd, although it should equal 0 . Therefore $e^{1905}=-1$, which gives $x_{0}^{2}=2^{k_{1005}} \leq 2^{1904}$, so the maximal value that $x_{0}$ can have is $2^{997}$. This value is attained in the case $x_{i}=2^{997-i}$ for $i=0, \ldots, 997$ and $x_{i}=2^{i-995}$ for $i=998, \ldots, 1995$.
Serond solution. First we show that there is an $n, 0 \leq n \leq 1995$, such that $x_{n}=1$. Suppose the contrary. Then each of $x_{n}$ belongs to one of the intervals $I_{-i-1}=\left[2^{-i-1}, 2^{-i}\right)$ or $I_{i}=\left(2^{i}, 2^{i+1}\right]$, where $i=0,1,2, \ldots$. Let $x_{n} \in I_{i_{n}}$. Note that by the formula for $x_{n}, i_{n}$ and $i_{n-1}$ are of different parity. Hence $i_{0}$ and $i_{1995}$ are also of different parity, contradicting $x_{0}=$ $x_{1506}$.
It follows that for some $n, x_{n}=1$. Now if $n \leq 997$, then $x_{0} \leq 2^{007}$, while if $n \geq 998$, we also have $x_{0}=x_{1995} \leq 2^{007}$.