SAUDI ARABIAN MATHEMATICAL COMPETITIONS

The answer is $292$.

We can see that if $x=pq$ for some integers $p,q>1$ and $\gcd (p,q)=1$ then $1<p,q<x$ which implies that $p\mid y$, $q\mid y$, then $pq\mid y$, contradiction.

Hence $x$ and $x+1$ must be the powers of primes. But one of these numbers is even so one of them must be the power of $2$. On the other hand, $n$ must be less than $2x$, otherwise $2x\mid y$ leads to $x\mid y$, contradiction. With the existence of $x$ we can easily choose $y$.

Thus, number $n$ satisfies the given condition if and only if there exists an exponent of $2$ less than $n$ and bigger than $n/2$, namely $x$ such that $x+1$ or $x-1$ is a power of some prime. We can check directly each range of numbers:

1. For each number $219\le n\le 255$ we can choose $x=127^1$, $x+1=2^7$.

2. For each number $257\le n\le 511$ we can choose $x=2^8$ and $x+1=257^1$.

3. For each number from $513\le n\le 1023$ we cannot choose any $x$ since $511$ and $513$ are not the powers of prime.

4. For each number from $1025\le n\le 2019$ we cannot choose any $x$ since $1023$ and $1025$ are not powers of prime.

Therefore, the total number of integers we need to find is $37+255=292$. $\square$