Chinese Girls' Mathematical Olympiad

Let $y_i = \frac{x_i}{2^{i/2}}$, for $i = 1, 2, \dots, 22$. It is well-known that $f(t) = t + \frac{1}{t}$ is decreasing on $(0, 1]$ and increasing on $[1, +\infty)$. For $1 \le i \le 11$, we have $\frac{1}{2^{12-i}} \le y_i \le \frac{1}{2^{11-i}}$, thus $y_i + \frac{1}{y_i} \le 2^{12-i} + \frac{1}{2^{12-i}}$. For $12 \le i \le 22$, we have $2^{i-12} \le y_i \le 2^{i-11}$, thus $y_i + \frac{1}{y_i} \le 2^{i-11} + \frac{1}{2^{i-11}}$. Hence, we have
$$\begin{aligned} \left(\sum_{i=1}^{22} x_i\right) \left(\sum_{i=1}^{22} \frac{1}{x_i}\right) &= \left(\sum_{i=1}^{22} y_i\right) \left(\sum_{i=1}^{22} \frac{1}{y_i}\right) \\ &\le \frac{1}{4} \left(\sum_{i=1}^{n} \left(y_i + \frac{1}{y_i}\right)\right)^2 \\ &\le \frac{1}{4} \left(\sum_{i=1}^{11} \left(2^{12-i} + \frac{1}{2^{12-i}}\right) + \sum_{i=12}^{22} \left(2^{i-11} + \frac{1}{2^{i-11}}\right)\right)^2 \\ &= \left(2^1 + 2^2 + \dots + 2^{11} + \frac{1}{2^1} + \frac{1}{2^2} + \dots + \frac{1}{2^{11}}\right)^2 \\ &= \left(2^{12} - 1 - \frac{1}{2^{11}}\right)^2. \end{aligned}$$
When
$$x_i = \begin{cases} 2^{i-1}, & 1 \le i \le 11, \\ 2^i, & 12 \le i \le 22, \end{cases}$$
equality holds. Thus, the maximum value sought is $(2^{12} - 1 - \frac{1}{2^{11}})^2$. $\square$