Are you sure this is all the conditions? I think a restriction for $y $ is needed

Is this correct? See that $x$ or $x+1$ can only be perfect powers of $2$, and one of them has to be prime. Otherwise, there exist numbers belonging to [1,x) such that they divide $x$ or $x+1$ which contradicts our claim. Since, $256$ is too low for our $n$, and for $1024$ both $1023$ and $1025$ are composite. Similarly, neither $511$ nor $513$ is a prime. So, there do not exist a $n$ such that $y$ is divisible by all natural numbers from $1$ to $n$.

Official wording from Romanian: Cˆate numere naturale n satisfac urm˘atoarele condit¸ii: i) 219 ≤ n ≤ 2019, ii) exist˘a x, y ∈ Z astfel ˆıncˆat 1 ≤ x < n < y ¸si y este divizibil cu toate numerele naturale de la 1 la n cu except¸ia numerelor x ¸si x + 1 cu care y nu este divizibil. (taken from pregatirematematicaolimpiadejuniori.wordpress.com) Google translate: How many natural numbers satisfy the following conditions:   i) 219 ≤ n ≤ 2019,   ii) there is x, y ∈ Z so that 1 ≤ x <n <y ¸ and y is divisible by all numbers   natural from 1 to n with the exception of the numbers x and x + 1 with which y is not divisible.