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 $219 \leq n \leq 255$ we can choose $x = 127^{1}$, $x+1 = 2^{7}$.

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

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

4. For each number from $1025 \leq n \leq 2019$ 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 $37 + 255 = 292$. $\square$