$n=1$ is obviously superprime. Assume now that $n \geq 2$. Let $1=d_1<d_2<\ldots<d_k=n$ be its divisors. Then, $d_2$ is a prime and $d_2-d_1=d_2-1$ is a prime, too. Thus, $d_2=3$. Note that $d_{k-1}d_2=n,$ and so $d_{k-1}=\dfrac{n}{3}$. Therefore, $\dfrac{2n}{3}=n-\dfrac{n}{3}=d_k-d_{k-1}$ is a prime, too, which implies that $\dfrac{2n}{3}=2,$ that is $n=3$, which evidently works.