3. Given the sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfies
$$
a_{n+2}=a_{n+1}-a_{n} \text {. }
$$

If the sum of the first 1000 terms of the sequence is 1000, then the sum of the first 2014 terms is $\qquad$ .
3. 1000 .

From $a_{n+2}=a_{n+1}-a_{n}$, we get
$$
a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n} \text {. }
$$

Therefore, for any positive integer $n$ we have
$$
a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0 \text {. }
$$

Also, $2014=1000(\bmod 6)$, so
$$
S_{2014}=S_{1000}=1000 \text {. }
$$