Brazilian Mathematical Olympiad

Suppose there are $15$ such integers, none prime. If we list the primes in order, the $15$th is $p_{15} = 47$. Now let $p(n)$ be the smallest prime dividing $n$. Take $N$ to be that integer $n$ among the $15$ which has the largest $p(n)$. Then since $N$ is not prime and $p(n)$ is its smallest prime factor, we must have $N \geq p(n)^2$. Since all $15$ integers are relatively prime, they must all have different $p(n)$s. Hence $p(n) = p_{15}$, so $N \geq p_{15}^2 > 1998$. Contradiction.