Problem 1031

The answer is affirmative. We prove it by inductive construction.
Take $a_{1}$ as a positive integer such that $2 a_{1} \notin \left\{b_{n}\right\}_{n=1}^{\infty}, a_{1}>b-a$ (for example, if $b_{n_{0}}>b-a+1$, take $a_{1}=b_{n_{0}}-1$). Assume that $a_{1}, a_{2}, \cdots, a_{k}$ have been chosen such that
$a_{i+1}-a_{i} \in \{a, b\}, a_{m}+a_{l} \notin \left\{b_{n}\right\}_{n=1}^{\prime \prime 1}(1 \leqslant m \leqslant k, 1 \leqslant l \leqslant k)$. Consider
$$\begin{array}{l} \text { (I) } a_{1}+a_{k}+a, a_{2}+a_{k}+a, \cdots, a_{k}+a_{k}+a, 2 a_{k}+2 a \text {; } \\ \text { (II) } a_{1}+a_{k}+b, a_{2}+a_{k}+b, \cdots, a_{k}+a_{k}+b, 2 a_{k}+2 b \text {. } \end{array}$$

Assume (I) contains a term $b_{u}$ from $\left\{b_{n}\right\}_{n=1}^{n}$, and (II) contains a term $b_{v}$ from $\left\{b_{n}\right\}_{n=1}^{n}$. Since
2 a_{k}+2 b0$. By the inductive hypothesis, $a_{i}-a_{j}=c a+d b, c, d$ are non-negative integers. Thus, $c a+d b=b-a$. Therefore, $d=0, b=(c+1) a$, which contradicts the fact that $a$ does not divide $b$. Case $2 \quad b_{u}=2 a_{k}+2 a$. In this case, $a_{k}-a_{j}=b-2 a$. By $1 \leqslant j \leqslant k$ and the inductive hypothesis, $a_{k}-a_{j}=c^{\prime} a+d^{\prime} b, c^{\prime}, d^{\prime}$ are non-negative integers. Thus, $c^{\prime} a+d^{\prime} b=b-2 a$. Therefore, $d^{\prime}=0, b=\left(c^{\prime}+1\right) a$, which is a contradiction. Hence, (I) or (II) does not contain any term from $\left\{b_{n}\right\}_{n=1}^{\infty}$. Therefore, we can take $a_{k+1}=a_{k}+a$ or $a_{k}+b$ such that
a_{k+1}+a_{i} \notin \left\{b^{\prime \prime}\right\}_{n=1}^{\prime \prime}, 1 \leqslant i \leqslant k+1 \text {. }
$$$$

The proposition is proved.