Dodging permutation sums Let $X =\left (x_1, x_2, \ldots, x_n \right)$ be a finite sequence of distinct positive integers, with $n \geq2$, and $S(X) = \left\{ x_1, x_1 + x_2, \ldots, \sum_{i=1}^{j}{x_i}, \ldots, \sum_{i=1}^{n}{x_i} \right\}$ the set of such partial sums of the sequence.
Given any set $Y = \left\{ y_1, y_2, \ldots, y_{n-1} \right\}$, where  $x_1 < y_k < \sum_{i=1}^{n}{x_i}$, is it possible to show that there always exists a permutation of the sequence $X$, called $X^\prime$, such that $S(X^\prime)\cap Y = \emptyset$?

A geometric statement/intuition of the problem could be this: let there be a line segment and a set $X$ of $n$ distinct lengths, totaling up to the segment length. Given a arbitrary set $Y$ of $n-1$ points on the open segment, does there always exist a permutation of $X$ so that the points it induces on the segment do not coincide with any of the $Y$ points?
 A: Yes, there always exists such a permutation.
We'll prove the generalization where $Y$ is only required to satisfy $|Y|\leq n-1$ and $0<y<\sum_{i=1}^n x_i$ for $y\in Y.$ The proposition is true when $n=1.$ We'll use induction, so assume the proposition is true for smaller $n,$ and we need to prove it for $n.$ If $Y$ is empty there's nothing to prove. Otherwise, there are three cases. In each case we apply the proposition with $n-1$ and $X^@,Y^@$ where $X^@=X\setminus\{\max X\}$ and $Y^@$ depends on the case.
1. $\min Y<\max X$ and $\max X\not\in Y$
In this case we remove the $y\in Y$ less than $\max X$ by taking $Y^@=\{y-\max X\mid y\in Y\text{ and }y>\max X\}.$ Prepend $\max X$ to the permutation of $X^@$ given by the induction hypothesis.
2. $\min Y\leq\max X$ and $\max X\in Y$
In this case we remove the $\max X$ from $Y$ and shift the elements of $Y$ down by $\max X$ when possible by taking $Y^@=\{y\in Y\mid Y<\max X\}\cup\{y-\max X\mid y\in Y\text{ and }y>\max X\}.$ Prepend $\max X$ to the permutation of $X^@$ given by the induction hypothesis. The one-element partial sum $\max X$ will hit $Y$ at $\max X,$ but we can fix this by swapping $\max X$ with the next element. The new first element cannot be in $Y^@,$ so cannot be in $Y.$
3. $\min Y>\max X$
In this case we remove $\min Y$ by taking $Y^@=\{y-\max X\mid y\in Y\setminus\{\min Y\}\}.$ Prepend $\max X$ to the permutation of $X^@$ given by the induction hypothesis. Some partial sum $\max X + x_1 + \dots + x_i$ (adopting the ordering given by induction for notation) may hit $\min Y,$ but we can fix this by swapping $\max X$ with $x_{i+1}.$
