. Let $x_{1}, \ldots, x_{n}$ be real numbers strictly greater than $1$. We count the number of reals of the form $\varepsilon_{1} x_{1}+\cdots+\varepsilon_{n} x_{n}, \varepsilon_{i} \in\{0,1\}$ that are in a given interval of length 1. What is the maximum number that can be obtained this way?
. Let's show that the maximum number of such reals is $\binom{n}{[n / 2]}$ where $[\cdot]$ denotes the floor function. We achieve this bound by taking, for example, $x_{i}=i+\frac{1}{2^{\ell}}$. In this case, the sums of $[n / 2]$ of the $x_{i}$ are all distinct (due to the uniqueness of the binary representation) and all within the interval $[[n / 2],[n / 2]+1]$. Therefore, they are exactly $\binom{n}{[n / 2]}$ in number. Let's now show that this value cannot be exceeded. First, note that if we have a permutation $\sigma$ of the set $\{1, \ldots, n\}$, then at most one of the sets $J_{i}^{\sigma}=\{\sigma(1), \ldots, \sigma(i)\}$ satisfies $\sum_{j \in J_{i}^{\sigma}} x_{j} \in I$, since the $x_{j}$ are greater than 1. The sum in the statement involves summing the $x_{j}$ over a subset $J$ of $\{1, \ldots, n\}$. If such a subset $J$ has cardinality $i$, there are $i!(n-i)!$ permutations $\sigma$ such that $J=J_{i}^{\sigma}$. Let $f(J)$ be the set of these permutations, and let $T$ be the set of $J \subset\{1, \ldots, n\}$ such that $\sum_{j \in J} x_{j} \in I$. According to the previous remark, the sets $f(J)$ for $J \in T$ are pairwise disjoint, and since $\operatorname{Card}(f(J))=\operatorname{Card}(J)!(n-\operatorname{Card}(J))!$, we have

$$
\sum_{J \in T} \operatorname{Card}(J)!(n-\operatorname{Card}(J))!\leqslant n!
$$

since the set of permutations of $\{1, \ldots, n\}$ has cardinality $n!$. Now, $t!(n-t)!$ is minimal for $t=[n / 2]$, so the sum on the left is bounded below by $\operatorname{Card}(T) \times[n / 2]!(n-[n / 2])!$. Dividing the resulting inequality by $n!$, we get

$$
\operatorname{Card}(T) \leqslant\binom{ n}{[n / 2]}
$$

which is the desired inequality.