4. Given the sequence $\left\{a_{n}\right\}(n \geqslant 1)$ satisfies $a_{n+2}=$ $a_{n+1}-a_{n}$ and $a_{2}=1$. If the sum of the first 1999 terms of this sequence is 2000, then the sum of the first 2000 terms equals ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002
4. (C).

From the given $a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}$ $=-a_{n}$, i.e., $a_{n+3}+a_{n}=0$, therefore $a_{n}+a_{n+1}+a_{n+2}+$ $a_{n+3}+a_{n+4}+a_{n+5}=0$, which means the sum of any 6 consecutive terms of the sequence is 0. Noting that $1999=6 \times 333+1,2000=6 \times 333+$ 2, we can conclude that the sum of the first 1999 terms is $a_{1}$, i.e., $a_{1}=2000$. Thus, the sum of the first 2000 terms is $a_{1}+a_{2}=2001$.