Explanation:

n² + 5n + 6
(n + 2) (n + 3)
(n + 2) and (n + 3) are two consecutive numbers.
One of (n + 2) or (n + 3) has to be even.
We need n such that (n + 2) (n + 3) is a multiple of 3.
(n + 2) or (n + 3) should be a multiple of 3.
n leaves a remainder of 0 or 1 when divided by 3.
n could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....
So, from among the first 99 natural numbers, N cannot take 2, 5, 8, 11, 14…….98.
There are 33 numbers in this list – we are effectively listing all numbers {3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2}.
So, N can take 99 – 33 = 66 values.