Four. (50 points) Given $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers, and $\left|x_{i}\right| \geqslant 1$, for $X=\{1,2, \cdots, n\}$, the subset $A=$ $\left\{i_{1}, i_{2}, \cdots, i_{t}\right\}$, define
$$
S(A)=x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{1}},
$$

where it is specified that $S(\varnothing)=0$. Question: From the $2^{n}$ such sums, what is the maximum number that can be selected so that the absolute value of the difference between any two is less than 1?
Let's assume all $x_{i}>0$.
In fact, if there is some $x_{i}<0$, then replace $x_{i}$ with $-x_{i}$, and change the set $A$ to:
$A^{\prime}=A \cup\{i\}(i \notin A)$ or $A^{\prime}=A \backslash\{i\}(i \in A)$.
Thus, the "sum $S(A)$" becomes $S\left(A^{\prime}\right)=S(A)-x_{i}$, meaning all $2^{n}$ sums increase by $-x_{i}$, and the difference between any two "sums" remains unchanged.
Therefore, $x_{i} \geqslant 1$.
Let $A_{1}, A_{2}, \cdots, A_{k}$ be the selected subsets of $X$, satisfying
$$
\left|S\left(A_{i}\right)-S\left(A_{j}\right)\right|<1 .
$$

Thus, each $A_{i}$ must not contain any other.
Otherwise, if $A_{i} \subseteq A_{j}$.
Then $\left|S\left(A_{i}\right)-S\left(A_{j}\right)\right|=\left|S\left(A_{i} \backslash A_{j}\right)\right| \geqslant 1$.
This leads to a contradiction.
By Sperner's Theorem, the number of selectable sets is
$$
k \leqslant \mathrm{C}_{n}^{\left[\frac{n}{2}\right]}.
$$

Furthermore, if $x_{i}=1$, then $X=\{1,2, \cdots, n\}$ has exactly $\mathrm{C}_{n}^{\left[\frac{n}{2}\right]}$ subsets of size $\left[\frac{n}{2}\right]$ that are mutually non-contained, and for each $A_{i}$, $S\left(A_{i}\right)=\left[\frac{n}{2}\right]$.
Thus, $\left|S\left(A_{i}\right)-S\left(A_{j}\right)\right|=0<1$.
Therefore, the maximum number of sets is $\mathrm{C}_{n}\left[\frac{n}{2}\right]$.
[Note] Sperner's Theorem
Let $|X|=n$, and $\mathscr{F}=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ be a family of mutually non-contained subsets of $X$. Then $k \leqslant \mathrm{C}_{n}^{\left[\frac{n}{2}\right]}$.