Problem 1343

1. Terminology and Lemma Introduction:
- We define a "k-chain, m narrow" as a sequence of integers $a_1, a_2, \ldots, a_k$ such that $0 < a_{j+1} - a_j \le m$ for $1 \le j \le k-1$.
- Lemma: If a set $S$ satisfies the property $P$ (the existence of a k-chain, m narrow for any $k$), and $S = A \cup B$ with $A$ and $B$ disjoint, then at least one of $A$ or $B$ must satisfy the property $P$.

2. Proof of the Lemma:
- Suppose $A$ does not satisfy the property $P$. This means there exists some $k_0$ such that $A$ does not have a k-chain, m narrow for $m = m_0$.
- Since $S$ satisfies $P$ for $m = m_0$, there exists a $k_0$-chain, $m_0$ narrow in $S$.
- To ensure $A$ does not have this property, we must have removed at least $k_0$ elements from this chain to form $B$.
- These $k_0$ removed elements form a $k_0$-chain, $m_0$ narrow in $B$.
- This argument holds for any $k$, hence $B$ satisfies $P$ for $m = k_0 m_0$.

3. Application of the Lemma to the Problem:
- We start with the set of all positive integers $\mathbb{N}$ and decompose it into $r$ disjoint subsets $A_1, A_2, \ldots, A_r$.
- By the lemma, if $A_1$ does not satisfy $P$, then the remaining set $S_1 = \mathbb{N} \setminus A_1$ must satisfy $P$.
- We continue this process: $S_1 = A_2 \cup S_2$, and if $A_2$ does not satisfy $P$, then $S_2$ must satisfy $P$.
- We repeat this until we reach $S_{r-2} = A_{r-1} \cup A_r$.
- By the lemma, at least one of $A_{r-1}$ or $A_r$ must satisfy $P$.

4. Conclusion:
- Since we have a finite number of steps and each step ensures that at least one subset satisfies $P$, we conclude that at least one of the subsets $A_i$ must satisfy the property $P$.

$\blacksquare$