jmc

For $n=1,$ $f(1)=1,$ so $$f(f(1))=f(1)=1.$$Thus, $n=1$ does not satisfy $f(f(n))=n+2.$ Henceforth, assume that $n\ge 2$.

Since $1$ and $n$ always divide $n$, we have that $f(n)\ge n+1$, so $f(f(n))\ge n+2$. Therefore, in order for $n$ to be superdeficient, $f(n)=n+1$ and $f(n+1)=n+2$. However, if $f(k)=k+1$, then $k$ must be prime. Therefore, we are searching for consecutive prime integers. However, one of those primes must necessarily be even, and the only even prime is $2$. Note that $f(2)=3$ and $f(3)=4$, so there is exactly $\boxed{1}$ superdeficient number: $2$.