Baltic Way 1990 · Problem 14

Solution:

Such numbers do exist. Let $M = 1990!$ and consider the sequence of numbers $1 + M, 1 + 2M, 1 + 3M, \ldots$ For any natural number $2 \leq k \leq 1990$, any sum of exactly $k$ of these numbers (not necessarily different) is divisible by $k$, and hence is a composite number. It remains to show that we can choose $1990$ numbers $a_{1}, \ldots, a_{1990}$ from this sequence which are relatively prime. Indeed, let $a_{1} = 1 + M$, $a_{2} = 1 + 2M$ and for $a_{1}, \ldots, a_{n}$ already chosen take $a_{n+1} = 1 + a_{1} \cdots a_{n} \cdot M$.