Non-Divisibility

Rewrite the expression in terms of the sum of first n squares; $n(n+1)(2n+1)=6(1^2+2^2+3^2+...+n^2)$

It's obvious that $n(n+1)(2n+1)$ is a multiple of $6$ , thus the only possible remainders when reduced by modulo $18$ are $6,12$ . So we got 2 cases with 2 remainders;

Case 1: $6(1^2+2^2+3^3+...+n^2)=18x+6$

simplifying becomes; $1^2+2^2+3^3+...+n^2=3x+1$

We want $1^2+2^2+3^3+...+n^2\equiv{1\mod{3}}$

Using the cycle of the reduced residues of $n^2\mod{3}$ , we get the maximum of $n=96$

Case 2: $6(1^2+2^2+3^3+...+n^2)=18y+12$

simplifying; $1^2+2^2+3^3+...+n^2=3y+2$

We want $1^2+2^2+3^3+...+n^2\equiv{2\mod{3}}$

and using reduced residues of $n^2\mod{3}$ again, we yield the maximum $n=97$

Thus $97$

Went by trial and error method :

Trick is that denominator is 18 = 9 * 2 & so the numerator can't be even & multiple of 9  at the same time.

99 is out of option and even 98 but 97 is divisible by 6 and not by 18.

Alternate solution :

The given equation is a modified form of sum of squares to 'N' terms = N(N+1)(2N+1)/6 which means for any    value of 'N' by default is divisible by 6. By checking 98,99 are divisible by 18 because of  '99' which is already a multiple of 9 and so 97 is the answer.