Problem 1884

For each integer $n \geq 1$, let $f(n)$ be the number of lists of different positive integers starting with $1$ and ending with $n$, in which each term except the last divides its successor. Prove that for each integer $N \geq 1$ there is an integer $n \geq 1$ such that $N$ divides $f(n)$.
(So $f(1) = 1$, $f(2) = 1$ and $f(6) = 3$.)