Take $x_j=2^{i-1}$. I claim that if $\sum_{j=1}^n c_j 2^{j-1} = \sum_{j=1}^n 2^{j-1}$ with $\sum_{j=1}^n c_j\leq n$, then $c_j=1$ for all $j$. The case $n=1$ is easy. In general note that $c_1$ is odd. We substract 1 on both sides of the equation and divide by 2, replace $c_2$ by $c_2+\frac{c_1-1}{2}$, and finally shift all indices by 1. Then we obtain an equation of the same form as before for a smaller value of $n$, and are done by induction. The same argument shows that ${1, 2, 4, 8, \ldots}$ is $i$-sum avoiding for every $i$, however, for larger $i$ there should be more efficient choices.