4. Given $a=\underset{2016 \uparrow}{55 \cdots 5}$. Then the remainder when $a$ is divided by 84 is ( ).
(A) 0
(B) 21
(C) 42
(D) 63
4. D.

Since $84=4 \times 3 \times 7$, and $4,3,7$ are pairwise coprime, we consider the remainders when $a$ is divided by $4,3,7$ respectively.

Notice that the last two digits of $a$ are $55 (55=4 \times 13 + 3)$, so $a$ leaves a remainder of 3 when divided by 4.

The sum of the digits of $a$ is $5 \times 2016 \div 3 = 3360$, so $3 \mid a$.
It is easy to see that $555555 = 7 \times 79365$,
$2016 \div 6 = 336$,
thus $7 \mid a$.
Since 3 and 7 are coprime, we have $21 \mid a$.
We list the numbers divisible by 21:
$$
21, 42, 63, 84, \cdots \text {. }
$$

Since 21 and 4 are coprime, we find the number less than 84 that leaves a remainder of 3 when divided by 4, which is 63. Therefore, the remainder when $a=\underset{2016 \uparrow}{55 \cdots 5}$ is divided by 84 is 63.