Let $n \geq 2$ such that $f(f(n))=2$. The largest proper divisor of $f(n)$ is 1 if and only if $f(n)$ is a prime number.

If $f(n)=2$, this is equivalent to $n$ being a prime number.

If $f(n)=p \geq 3$, then $p-1$, the largest proper divisor of $n$ is an even number. This means that $n$ is even and therefore, its largest proper divisor is $\frac{n}{2}$. This is equivalent to $n=2(p-1)$.

Hence, the positive integers $n$ such that $f(f(n))=2$ are prime numbers or numbers $n=2(p-1)$ for any prime number $p$.