Solution:

The answer is (D). If $n$ has four positive divisors, they are, in increasing order, $1, p, q, n$ with $p$ a prime number.
If $q$ is a prime number, we can conclude that $n = p \cdot q$, because $n$ has no other divisors. The sum of the positive divisors is $1 + p + q + p \cdot q = (1 + p) + q(1 + p) = (1 + p)(1 + q)$ which by hypothesis must equal $42 = 2 \cdot 3 \cdot 7$. Neither $(1 + p)$ nor $(1 + q)$ can equal 1, nor can either equal 42, otherwise the other would equal 1. Since 42 has only three prime factors, one of them equals $(1 + p)$ or $(1 + q)$, and hence must still be prime when decreased by one. The only possibility is therefore that $p + 1 = 3$ and $q + 1 = 2 \cdot 7 = 14$. Since $p = 3 - 1 = 2$ and $q = 14 - 1 = 13$ are prime, $n = 2 \cdot 13 = 26$ has the required properties.

Let us now consider the other possibility, that $q$ is not a prime number. Then $q$ has as divisors only $1, p$ and itself, otherwise other divisors would also divide $n$. Thus $q = p^{2}$ and $n = p^{3}$. The sum of the positive divisors is $1 + p + p^{2} + p^{3} = 1 + p(1 + p + p^{2}) = 42$, so $p(1 + p + p^{2}) = 41$. But, since 41 is a prime number, in this case there are no solutions.