See
Different subset sums of a subset of naturals
Hardest problem from Sankt-Petersburg 239MO olympiad 2007
Let $\{s_1,\cdots,s_{2^n}\} = \bigg\{ \sum_{i\in I} x_i \bigg| I \subset\{1,\cdots,n\} \bigg\}$, $s_1\leq \cdots\leq s_{2^n}$. By the given condition, $s_k - s_l \geq |k-l|$ for every $1\leq k,l\leq 2^n$. By simple calculations, we get the desired result.
\begin{align*}
& \sum_{k=1}^{2^n} s_k = 2^{n-1} \sum_{i=1}^n x_i \\
& \sum_{k=1}^{2^n} s_k^2 = 2^{n-1} \sum_{i=1}^n x_i^2 + 2^{n-1} \sum_{1\leq i<j\leq n} x_i x_j \\
\Longrightarrow & \sum_{i=1}^n x_i^2 = 2\cdot \frac{\sum\limits_{k=1}^{2^n} s_k^2}{2^{n-1}} - \bigg( \frac{\sum\limits_{k=1}^{2^n} s_k}{2^{n-1}} \bigg)^2 \\
\Longrightarrow & 2^{2n-1} \sum_{i=1}^n x_i^2 = 2^{n} \sum_{k=1}^{2^n} s_k^2 - \bigg( \sum_{k=1}^{2^n} s_k \bigg)^2 = \frac{1}{2}\sum_{k,l=1}^{2^n} (s_k-s_l)^2 \\
& \qquad\qquad\quad\; \geq \frac{1}{2} \sum_{k,l=1}^{2^n} (k-l)^2 = 2^n \sum_{k=1}^{2^n} k^2 - \bigg( \sum_{k=1}^{2^n} k \bigg)^2 = 2^{2n-1} \cdot \frac{4^n -1}{3}
\end{align*}