Clearly if $a$ fails, then all $b$ multiples of $a$ will also fail. We will prove that all primes $p$ except for $2,3$ fail. Clearly the question is equivalent to asking if for every $0\le k \le p-1$, exists $0 \le x \le p-1$ such that $\frac{x(x+1)(2x+1)}{6}\equiv k \mod{p}$. This is equivalent to every $k$ having $x$ s.t. $$x(x+1)(2x+1)\equiv k \mod{p}$$since $p$ and $6$ are coprime. Now this means $x(x+1)(2x+1)$ for every $0\leq x \leq p-1$ must have distinct residues mod $p$. Yet this cannot be true as $0$ and $p-1$ are both $0 \mod{p}$. Hence every $N$ that would work must be of the form $2^a 3^b$. Now we will prove all these solutions are indeed solutions. Notice that we may use Chinese Remainder Theorem as the remainder of $x(x+1)(2x+1)$ mod $2^a$ depends only on $x\mod{2^{a+1}}$. This is similar for $3^b$. So we need only show $2^a$ works for all $a$ and that $3^b$ works for all $b$. We shall only show for $2^a$ as the case for $3^b$ is similar. Now, let us proceed by induction. First define $f(n)=1^2+2^2+...+n^2$ We will prove that for all $n$, for all $k$, exists integer $x$ satisfying $0\le x\le 2^{n+1}-1$ such that $f(x) \equiv k \mod{2^n}$. It is obvious this is equivalent to the original statement. The base case of $n=1$ can be very easily verified. Now if $n$ works, then notice that $f(2^{n+1}) \equiv 2^n \mod{2^{n+1}}$. This is because $$v_2(f(2^{n+1}))=n$$ which can be easily checked. Now we use the induction hypothesis. Suppose by contradiction that $k$ cannot be achieved mod $2^{n+1}$. Then let $a$ be a pos. integer less than $2^{n+1}$ such that $f(a)\equiv k \mod{2^n}$. Now since $k$ is unachievable, then that must mean $f(a)\equiv k+2^n \mod{2^{n+1}}$. But then consider $f(a+2^n)\equiv f(a)+2^n \equiv k \mod{2^{n+1}}$ so contradiction and we are done.