We call two terms $ a_i$ and $ a_j$ friends if $ a_i|a_j$ or $ a_j|a_i$ and enemies otherwise.We call a sequence "good" if any two terms are friends. Assume there is no good infinite sequence. Let $ \displaystyle a_1=a{}_1{}_1, a{}_1{}_2,\dots a{}_1{}_n$ be the longest good sequence containing $ a_1$, where $ 1_1, 1_2, \dots 1_n$ are indices.This means that for any $ \displaystyle i$ distinct from $ \displaystyle \{1_1, 1_2, \dots 1_n\}$, there exists $ j\leq n$ such that $ a_i$ and $ a{}_1{}_j$ are enemies. It's clear that there exists a number $ a{}_1{}_k$ with an infinite number of enemies. Denote $ a{}_1{}_k=b_1.$Therefore there exists an infinite set $ I_1$ of indices such that for any $ i\in I_1$, $ a_i$ and $ b_1$ are enemies. Assume that we have constructed the numbers $ b_1, b_2\dots b_n$ such that any two are enemies and there exists an infinite set $ I_n$ of indices such that for any $ b_i$ and for any $ j\in I_n$, $ b_j$ and $ a_i$ are enemies. Pick a number $ x$ from $ I_n$ such that $ a_x$ is distinct from $ b_i$, for any $ i\leq n$, and let $ a_x=a{}_n{}_1, a{}_n{}_2, \dots a{}_n{}_m$ be the longest good sequence containing $ a_x$, with indices $ a{}_n{}_i\in I_n$, for any $ i\leq m$. Therefore any element $ y$ of $ I_n$, which is not from the set $ \{n_1, n_2\dots n_m\}$, has the property that $ a_y$ is an enemy with some $ a{}_n{}_j$. Since the set $ I_n -\{n_1, n_2\dots n_m\}$ is infinite, there exists an indice $ n_k$ which is an enemy with an infinite number of elements with indices from $ I_n$.. Let $ a{}_n{}_k=b_{n+1}$ and let $ I_{n+1}\subset I_n$ be the set of indices with which $ b_{n+1}$ is an enemy. Therefore we have constructed the number $ b_{n+1}$ which is an enemy with $ b_1, b_2\dots b_n$, and the infinite set of indices $ I_{n+1}$ such that for any $ y\in I_{n+1}$, $ a_y$ is an enemy with any $ b_{i}$, $ \forall$ $ i\leq n+1$. Thus, by induction, we have constructed an infinite subsequence $ \{b_n\}$ of enemies.

The problem is just a nice application of Dillworth theorem,which is similar with Romania 2005 TST Day 1 problem 2