Find all integers n such that S(n) = S(2n) = ... = S(n*n), where S is the sum of the base-10 digits.

Well, I'm always willing to go for the easy answer.

The obvious answer is 9.

S(9)=S(18)=S(27)=S(36)=S(45)=S(54)=S(63)=S(72)=S(81)

Arguably 0 and 1 work also, but I think the form of the problem implies that N >=2.

If there are any larger answers, they must obviously be a multiple of 9, but 18 doesn't work, so I'm going to stop looking.