8・165 Given a sequence of natural numbers $\left\{x_{n}\right\}$ that satisfies
$$x_{1}=a, x_{2}=b, x_{n+2}=x_{n}+x_{n+1}, n=1,2,3, \cdots$$

If one of the terms in the sequence is 1000, what is the smallest possible value of $a+b$?
[Solution] Take the sequence $\left\{t_{n}\right\}$ satisfying:
$$t_{1}=1, t_{2}=0, t_{n+2}=t_{n}+t_{n+1}, n=1,2,3, \cdots$$

It is easy to prove that for any natural number $n$ we have
$$x_{n}=t_{n} a+t_{n+1} b$$

The terms of the sequence $\left\{t_{n}\right\}$ that are less than 1000 are $1,0,1,1,2,3,5,8,13,21,34,55,89,144$, 233,377,610,987. It is easy to prove that the equations
$$\begin{array}{l}
610 a+987 b=1000,377 a+610 b=1000 \\
233 a+377 b=1000,144 a+233 b=1000
\end{array}$$

all have no positive integer solutions. However, the equation $89 a+144 b=1000$ has a unique positive integer solution $a=8$, $b=2$. Since $89=t_{13}$, the equation $t_{n} a+t_{n+1} b=1000$ can only have positive integer solutions $a, b$ when $n \leqslant 13$. If there is a natural number $k$ such that $t_{k} a^{\prime}+t_{k+1} b^{\prime}=1000=8 t_{13}+2 t_{14} \\
& >10 t_{13} \geqslant 10 t_{k+1}
\end{aligned}$$

then $a^{\prime}+b^{\prime}>10$. Therefore, the minimum value of $a+b$ is 10.