4. A.

Notice that, $N^{2}=\left(n^{2}+10 n+12\right)^{2}-144$.
And $n \geqslant 2018$, thus,
$$
\left(n^{2}+10 n+11\right)^{2}<N^{2}<\left(n^{2}+10 n+12\right)^{2} \text {. }
$$

Therefore, $[N]=n^{2}+10 n+11$.
Since $2018 \equiv 2(\bmod 7)$, by checking one by one, we find that only when $n=2018$ meets the condition.
Thus, $1<a<\frac{5}{3},-\frac{2}{3}<s=1-a<0$.
4. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer,
$$
N=\sqrt{n(n+4)(n+6)(n+10)},
$$

and $[N]$ is divisible by 7, then the possible value of $n$ is ( ).
(A) 2018
(B) 2019
(C) 2020
(D) 2021