Ukrainian National Mathematical Olympiad

In the case $n = pq$, $S = 1 + p + q + pq = (1+p)(1+q)$. Suppose $p < q$.
Let us consider possible cases.
If $S = 40$, then $(p+1)(q+1) = 40$. Since $p+1 \ge 3$, the following cases are possible:
$$p+1=4, q+1=10 \Rightarrow p=3, q=9 \Rightarrow q \text{ is not prime.}$$
$$p+1=5, q+1=8 \Rightarrow p=4, q=7 \Rightarrow p \text{ is not prime.}$$
If $S = 41$, then $(p+1)(q+1) = 41$, since $41$ is prime, there are no solutions.
If $S = 42$, then $(p+1)(q+1) = 42$. The options are:
$$p+1=3, q+1=14 \Rightarrow p=2, q=13 \Rightarrow n=26.$$
$$p+1=6, q+1=7 \Rightarrow p=5, q=6 \Rightarrow q \text{ is not prime.}$$
So, in this case, we have a single number $n=26$, which satisfies the conditions.
In case $n = p^3$, $S = 1 + p + p^2 + p^3$. Note that if $p=2$, then $1 + p + p^2 + p^3 = 15$; if $p=3$, $1 + p + p^2 + p^3 = 40$; and if $p \ge 4$, $1 + p + p^2 + p^3 \ge 85$.
In this case, there is also a single number $n=27$, which satisfies the conditions.