2. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}(n \geqslant 1)$ satisfy
$$
a_{n+1}=2 b_{n}-a_{n}, b_{n+1}=2 a_{n}-b_{n}
$$
$(n=1,2, \cdots)$. If $a_{1}=2007, a_{n}>0(n=2,3$, $\cdots$ ), then $b_{1}$ equals $\qquad$ .
2.2007.

Since $b_{n}=\frac{1}{2}\left(a_{n}+a_{n+1}\right)$, we have
$$
a_{n+2}+2 a_{n+1}-3 a_{n}=0 \text{. }
$$

By the method of characteristic roots, we get
$$
a_{n}=\frac{2007+b_{1}}{2}+\frac{b_{1}-2007}{6}(-3)^{n} \text{. }
$$

If for any positive integer $n$, we have $a_{n}>0$, then it must be that $b_{1}=2007$.