Not sure if this is the most efficient way but here goes...

1 - insufficient. Take n = 1 and n = 3 for example.
They yield remainders of 0 and 8 resp. No unique value. So rule out 1.

2 - insufficient. Take n = 1 and n = 2.
They yield remainders of 0 and 3 resp. No unique value. So, rule out 2

A number that is not divisible by 2 AND not divisible by 3 can be
written as n = 6x + 1 or n = 6x - 1

For n = 6x+1
(n-1)*(n+1) / 24
= 6x * (6x+2) / 24
= 12x * (x+1) / 24

We know that x*(x+1) will be even or 0, so the above equation will
always yield a remainder of 0.

Similarly for n=6x-1
(n-1)*(n+1) / 24
= (6x - 2)*6x
= 12x*(x-1) / 24

Again, x*x(-1) will be even or 0 and hence remainder r will be 0.

So, C looks like the correct answer.

On the actual GMAT, in the interest of time, it might be worthwhile
trying n = 1, 3, 5, 7, 9 etc. We can see that the remainder is 8 when n
is not divisible by 3 and remainder is 0 when n is divisible by 3 (for
all values n > 0). Using this pattern, I guess we can settle on C.

thank you,
I would go for picking numbers on the real GMAT too.

jayhawk2001, that is a wonderful explanation! Thank you. Gives us a valuable technique, we can use for similar questions.