Proving sum of negative powers of 2 equals one before becoming greater than one Let $x$: $x_1 \geq x_2 \geq x_3\geq...\geq x_n$ be negative powers of 2 with sum greater than one. Then $\exists$ $l$ s.t. $x_1 + x_2 + ... + x_l = 1$. It seems obvious from examples, but I'm finding it a bit difficult to prove it. Any help would be highly appreciated.

Your comment is close.
Let $S_i = x_1+x_2+\cdots + x_i$.
For the first case $l=1$, the first sum $S_1 = x_1<1$. For the last case $l=n$, $S_n > 1$ (as given). For the increasing sum $S_l$, there must be a largest $l$ that satisfy $S_l < 1$.
Assume that $k$ is the greatest integer that satisfy $$S_k = \sum_{i=1}^kx_i < 1$$
Then the goal is to prove that the next sum
$$S_{k+1}=\sum_{i=1}^{k+1}x_i = 1.$$
As you pointed out, $S_k = \sum_{i=1}^k x_i$ is an integer multiple of $x_{k+1}$, so let $S_k = Ax_{k+1}$. Then from the hypothesis,
$$\begin{align*}
Ax_{k+1} &< 1\\
A &< x_{k+1}^{-1}
\end{align*}$$
$x_{k+1}^{-1}$ is an integer for being a positive power of $2$, and since both sides are integers,
$$\begin{align*}
A + 1 &\le x_{k+1}^{-1}\\
Ax_{k+1} + x_{k+1} &\le 1\\
S_{k+1} &\le 1
\end{align*}$$
Also by the definition of $k$, $S_{k+1} \not< 1$.
Then the only possibility is that $$S_{k+1} = \sum_{i=1}^{k+1}x_i = 1$$
