12.5 The sequence of natural numbers $\left\{x_{n}\right\}$ is constructed according to the following rule:
$$
x_{1}=a, \quad x_{2}=b, \quad x_{n+2}=x_{n}+x_{n+1}, \quad n \geqslant 1 .
$$

It is known that one of the terms in the sequence is 1000. What is the smallest possible value of the sum $a+b$?
(Recommended by the Soviet Ministry of Education, 1990)
［Solution］We can find the first few terms of the sequence:
$$
\begin{array}{l}
x_{1}=a, \quad x_{2}=b, \quad x_{3}=a+b, \quad x_{4}=a+2 b, \\
x_{5}=2 a+3 b, \quad x_{6}=3 a+5 b, \quad x_{7}=5 a+8 b, \cdots
\end{array}
$$

In general, we have the relation
$$
x_{n}=t_{n} a+t_{n+1} b
$$

where $t_{1}=1, \quad t_{2}=0, \quad t_{n+2}=t_{n}+t_{n+1}$.
This can be easily proven by mathematical induction.
It can be seen that, starting from $n=4$, the sequence $\left\{t_{n}\right\}$ is strictly increasing.
Therefore, if $k>n \geqslant 4$, and
$$
\begin{aligned}
1000 & =t_{n} a+t_{n+1} b \\
& =t_{k} a^{\prime}+t_{k+1} b^{\prime},
\end{aligned}
$$

then it must be true that
$$
\begin{aligned}
(a+b) t_{n+1} & >t_{n} a+t_{n+1} b \\
& =t_{k} a^{\prime}+t_{k+1} b^{\prime} \\
& >t_{k}\left(a^{\prime}+b^{\prime}\right) \\
& \geqslant t_{n+1}\left(a^{\prime}+b^{\prime}\right),
\end{aligned}
$$

which implies
$$
a+b>a^{\prime}+b^{\prime} \text {. }
$$

This indicates that to find the minimum value of $a+b$, we need to determine the largest $n$ for which the equation
$$
t_{n} a+t_{n+1} b=1000
$$

has positive integer solutions, and among the solutions, we need to select the one that minimizes the sum $a+b$.
The first few terms of the sequence $\left\{t_{n}\right\}$ are
$$
1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987, \cdots
$$

Clearly, the equation $610 a+987 b=1000$ and all subsequent corresponding equations cannot have positive integer solutions. This is because the left side of the equation is greater than 1000 for positive integers $a$ and $b$.
It can be verified that the equations
$$
\begin{array}{l}
377 a+610 b=1000, \\
233 a+377 b=1000, \\
144 a+233 b=1000,
\end{array}
$$

do not have positive integer solutions.
However, the equation
$$
89 a+144 b=1000
$$

has a unique positive integer solution:
$$
a=8, \quad b=2 .
$$

Thus, the minimum value of $a+b$ is 10.