A number has four factor is two cases: it's a prime number raised to the third power, so \(N=n^3\).
So in this case the factors are \(n^3\), \(n^2\), \(n\) and \(1\). Their sum is 31, so \(n^2+n+1=31\), \(n=\)5 and \(n^3=125\) or n=-6, but the natural numbers are positive so this is not valid. ( and 6 is not prime )

The other valid case is the one in which the number \(N=a*b\), where a and b are prime.
In this case the factors are \(1,a,b,a*b\), their sum is \(a+b+1=31\) or \(a+b=30\), where a and B are prime.
This sum is valid if the numbers \(a,b\) are \(7,23\) or \(13,17\) or \(11,19\). (please note that 1 and 29 is not a valid option)
So N can be \(7*23=161\) or \(13*17=221\) or \(11*19=209\).

Total values 4