Question 43
Let $n \in \mathbb{N}^*$. Show that for any choice of $x_1, x_2, \dots, x_n \in \mathbb{N}^*$ such that $x_k \le k$ for all $k \in \{1, 2, \dots, n\}$ and the sum $x_1 + x_2 + \dots + x_n$ is odd, there exist $\varepsilon_1, \varepsilon_2, \dots, \varepsilon_n \in \{-1, 1\}$ such that $\sum_{k=1}^n \varepsilon_k \cdot x_k = 1$.