If $n=3k+2$ then$3|S(n)+1$ but 3 doesn't divide n,so the answer is 2(pick 231 and 232)

39 and 40 do also work

Since the problem just mentioned about $d(n)$ , there is very little we could exploit. So ,we should look at an obvious property about $mod 3$, and then we see how the consecutive condition comes to play. Now,we arrive at a contradiction for $n\equiv 2 (\mod 2)$ within a minute. We should be able to find a construction shortly as w ejust examine 1 or 2 out of 3 numbers.