19.4.7 $\star \star$ The sequence of natural numbers $\left\{x_{n}\right\}$ is constructed according to the following rule:
$$
x_{1}=a, x_{2}=b, x_{n+2}=x_{n}+x_{n+1}, n \geqslant 1 .
$$

It is known that one of the terms in the sequence is 1000. What is the smallest possible value of the sum $a+b$?
It is easy to verify that the terms in the sequence satisfy the relation $x_{n}=t_{n} a+t_{n+1} b$, where $t_{1}=1$, $t_{2}=0$, and $t_{n+2}=t_{n}+t_{n+1}$.
The sequence $\left\{t_{n}\right\}$ is essentially the Fibonacci sequence, with the first few terms being:
$$
1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987, \cdots \text {. }
$$

Clearly, the equations $610 a+987 b=1000$ and subsequent equations formed by later numbers in the sequence cannot have positive integer solutions.
After practical attempts, it is not difficult to find that the equations $377 a+610 b=1000$, $233 a+377 b=1000$, and $144 a+233 b=1000$ do not have positive integer solutions, while the equation $89 a+111 b=1000$ has a unique solution $a=8, b=2$.
On the other hand, if
$$
t_{n} a+t_{n+1} b=t_{k} a^{\prime}+t_{k+1} b^{\prime}=1000,
$$

and $kt_{k} a^{\prime}+t_{k+1} b^{\prime}=t_{n} a+t_{n+1} b>t_{n}(a+$ b), then $a^{\prime}+b^{\prime}>a+b$. Therefore, $a+b=8+2=10$ is the smallest possible value.