1987
Problem - 179
Find $n$ different positive integers such that any two of them are relatively prime, but the sum of any $k$ ($k < n$) of them is a composite number.
The $n$ number $a_i = i \cdot n! +1$, ($i=1, 2, \cdots, n$) satisfy the requirement.
First, the sum of any $k$ of them is $(m\cdot n! + k)$ for some integer $m$. This number clearly is a multiple of $k$ (because $k < n \implies k|n!$)
Secondly, if $a_i=i\cdot n! + 1$ and $a_j = j\cdot n! +1$ ($i>j$) have a common primer factor $p$, then $p | (i-j)\cdot n!$. Because $0< i - j < n$, this implies $p |n!$. This cannot hold because $i\cdot n! + 1$ and $n!$ are relatively prime.