Let $x_1,\cdots , x_n$ be a sequence of real number such that $x_i\geq 1$ for all $1\leq i\leq n$, $S=\{\alpha_1x_1+\cdots +\alpha_nx_n  \alpha_i\in\{0,+1,1\}\}$ and $I=[a,b)$ be a Interval with length $2$. So I was wondering if there was any subsequent upper bound on $I \cap S$. Is there a general bound when $2$ is replaced by a positive real $\alpha$? Thanks in advance for your answers and comments. I haven't been able to guess anything, but this bound on size of a set reminded me of Sperner's Theorem but not sure about it.

For a given $A\subset\{1,\dots,n\}$, let $S_A$ denote the multiset of $\alpha_1 x_1+\cdots+\alpha_n x_n$ with $\alpha_i=\pm 1$ for $i\in A$ and $\alpha_i=0$ for $i\not\in A$. Note that, as multisets, $$ S=\bigcup_{A\subset\{1,\dots,n\}} S_A,$$ so that $$ I\cap S=\bigcup_{A\subset\{1,\dots,n\}} (I\cap S_A).$$ By the Sperner bound, realized in this problem by Erdős (1945), we have $$ I\cap S_A\leq {A \choose \lfloor A/2\rfloor}, $$ with equality for $I=[1,1)$ and $x_i=1$ for $i\in A$. Hence, $$ I\cap S\leq \sum_{A\subset\{1,\dots,n\}}{A \choose \lfloor A/2\rfloor}=\sum_{m=0}^n {n\choose m}{m \choose \lfloor m/2\rfloor}, $$ with equality for $I=[1,1)$ and $x_1=\dots =x_n=1$. 


Expanding on my comment and continuing GH from MO's answer: For $\alpha >2$, let $r:=\lceil\alpha/2\rceil+1$. Since a halfopen interval $I$ of length $\alpha$ cannot contain $r$ numbers whose pairwise distance is at least $2$, for any fixed $A \subseteq \{1, \dots, n\}$, the set $A_+:=\{i \in A; \alpha_i = +1\}$ cannot contain an inclusion chain of length $r$. By Erdos's generalization of the Sperner Theorem, we have $$\lvert I\cap S_A \rvert \le \sum_{k=\lfloor A/2\rfloor\lfloor (r1)/2\rfloor}^{\lfloor A/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k}, $$ with equality for $I=[r+1, r+1+\alpha)$ and $x_i=1$ for $i\in A$. Hence, $$I\cap S\leq \sum_{A\subset\{1,\dots,n\}}\sum_{k=\lfloor A/2\rfloor\lfloor (r1)/2\rfloor}^{\lfloor A/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k} = \sum_{m=0}^n {n\choose m}\sum_{k=\lfloor m/2\rfloor\lfloor (r1)/2\rfloor}^{\lfloor m/2\rfloor+\lfloor r/2\rfloor} {m \choose k}, $$ with equality for $I=[r+1, r+1+\alpha)$ and $x_i=1$ for all $1\le i\le n$. 

