Problem 1047

Given numbers in increasing order $a_{1} < a_{2} < \ldots < a_{n-1} < a_{n}$, where $a_{1} > 0$.

The number $a_{n}$, when paired with any other number, will not have their sum appear among the other numbers, since $a_{n}$ is the largest. If for any $1 \leq i \leq n-1$, $\left|a_{n}-a_{i}\right|=a_{n}-a_{i}$ does not appear among $a_{1}, a_{2}, \ldots, a_{i-1}, \ldots, a_{n-1}$, then we are done. If for every $i$ there exists some $1 \leq j \leq n-1$ such that $a_{n}-a_{i}=a_{j}$ and $i \neq j$, then all numbers except $a_{n}$ can be paired such that the sum of each pair is exactly $a_{n}$. Since such a pairing is only possible if $n-1$ is even, the statement of the problem is proven for cases where $n-1$ is odd, i.e., when $n$ is even. The only cases left from those where $n$ is odd are those where $a_{1}, a_{2}, \ldots, a_{n-1}$ can be paired, i.e., when

$$a_{k}+a_{n-k}=a_{n} \quad \text{for all} \quad 1 \leq k \leq n-1.$$

Let's examine the numbers $a_{n-1}$ and $a_{n-2}$. Their difference, according to (1), is

$$a_{n-1}-a_{n-2}=\left(a_{n}-a_{1}\right)-\left(a_{n}-a_{2}\right)=a_{2}-a_{1}.$$

Thus, $a_{n-1}-a_{n-2}=a_{2}-a_{1}$. If we choose $a_{n-1}$ and $a_{k}$, where $1 < k < n-1$, then $a_{n-1}+a_{k}>a_{n-1}+a_{1}=a_{n}$, meaning that neither the sum nor the difference of the found numbers is among the remaining ones. This proves the statement for odd $n$ as well.

Remarks. 1. The statement is not true for $n=3$, as the number triplet 1, 2, 3 demonstrates.

2. We did not use the fact that the $a_{i}$ numbers are integers, so the statement of the problem is true for any $n$-tuple of distinct non-negative numbers.