2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$

If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ .
2. 0 .

From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\begin{array}{l}
x_{1}=x_{1997}=a x_{1006}+b x_{1905}, \\
x_{2}=x_{1998}=a x_{1997}+b x_{1996}=u x_{1}+b x_{1995}, \\
x_{3}=a x_{2}+b x_{1}, \\
x_{4}=a x_{3}+b x_{2}, \\
\cdots \cdots \\
x_{1096}=a x_{1995}+b x_{1994} .
\end{array}
$$

Summing up, we get $\quad \sum_{i=1}^{1004} x_{i}=a \sum_{i=1}^{1006} x_{i}+b \sum_{i=1}^{1096} x_{i}$,
which simplifies to $(a+b-1) \sum_{i=1}^{1098} x_{i}=0$.
Since $a+b-1 \neq 0$, it follows that $\sum^{1096} x_{i}=0$.