Jeffrey Samuelson twitted a question to which I am happy to provide an answer, or two.
Since the question is about the properties of a product modulo 6, it seems proper to investigate the dependency of the product on the residues modulo 6. Let's form a table
The first number in a cell is the remainder of division by 6 of the number in the caption of a column. The number in parentheses - if present - shows a factor of that number. Clearely, in every row, the product of factors is at least 6. So that, for every possible value of N modulo 6, the product N(N+1)(2N+1) is divisible by 6.
An effort of putting together a table could have been spared with an observation that, for every integer N,
In words, a product of any three consecutive integers is divisible by 6. This is a somewhat simpler variant of the problem at hand. Why a variant? Are not problems rather different? Yes and no. Consider the difference
N(N+1)(2N+1) - N(N+1)(N+2) = N(N+1)(N-1) = (N-1)N(N+1).
Or, separating N(N+1)(2N+1),
N(N+1)(2N+1) = N(N+1)(N-1) = (N-1)N(N+1) + N(N+1)(N+2),
meaning that N(N+1)(2N+1) is the sum of two products of three consecutive integers, each of which is divisible by 6.
Finally, N(N+1)(2N+1)/6 is the sum of the first N squares: 1² + 2² + ... + N² (see, Jeff's derivation. This is called a combinatorial argument. Since N(N+1)(2N+1)/6 is thus an integer, 6 is bound to be a factor of N(N+1)(2N+1).