4. A sequence of numbers
$$
1,1,2,3,5,8,13,21,34,55,89,144,233, \cdots
$$

is constructed such that the first and second numbers are both 1, and from the third number onwards, each number is the sum of the two preceding numbers. What is the remainder when the 2014th number in this sequence is divided by 7? $(\quad)$.
(A) 0
(B) 2
(C) 4
(D) 6
4. D.

Let the sequence of numbers be $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, with the rule $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
Consider the sequence of remainders when these numbers are divided by 7:
$$
1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,0,1,1 \cdots \text {. }
$$

Observing that the 17th and 18th numbers in the remainder sequence are the same as the 1st and 2nd numbers, the sequence repeats.
Noting that, $2014=125 \times 16+14$.
Thus, the result we seek is 6.