here is how i read this problem.

First thing i notice from the problem stem is that 4 is an even number.... second thing i notice is that N^3-N is = 3 consecutive integers multiplied by each other..

From there you can use rules for odds and evens to determine if 4 is divisible.

Good rules to know
(even)*(even) = divisible by 4
(even)*(even)*(even) = divisible by 8

So we know from problem stem we have (n-1)*(n)*(n+1) or three consecutive integers... Well if n is odd then we know that other two numbers are even therefore divisible by 4

S1 says N = 2K + 1
2 * any number makes an even number... so even + 1 = odd number so N = odd statement is suff

S2 says that n^2 + n
not 100% sure but i think this is insufficient because we cant determine if n is odd or even etc.. or at least i cant determine experts can chime in..

(n)(n+1)/6 means that there are factors of 2 and 3 which doesn't really tell us if n is even or odd

my answer would be A