Call a set $A \subset {\mathbb N}$ good if it has the desired property, i.e. there is an $m$ such that for every $k$ there are $a_{1},a_{2},...,a_{k} \in A$ s.t. $0<a_{i+1}-a_{i} \leq m$, for $i=1,2,...,k-1$. Call a sequence $x$-bounded if it is strictly increasing and the difference between any two consecutive numbers is at most $x$. Lemma. I'll prove that if $A,B \subset {\mathbb N}$ such that $A \cap B =\emptyset$ and $A \cup B$ is good, then at least one of $A$ and $B$ is good. For any $k$, take an $m$-bounded sequnce $c_{1},c_{2},...,c_{k} \in A \cup B$ and let $l_{k}$ be the largest number for which there is a $j$ such that all $c_{j}$, $c_{j+1}$, ..., $c_{j+l_{k}-1}$ belong to the same set ($A$ or $B$). If the sequence $l_{k}$ is unbounded, then we are done, so suppose it is bounded and take $L>l_{k},\forall k$. In any sequence $m$-bounded sequence from $A \cup B$, the two subsequences formed from numbers belonging to the same set are $(L+1)m$-bounded, so at least one set must contain arbitrarily long $(L+1)m$-bounded sequences. Back to the problem, let $A_{1} \cup A_{2} \cup ... \cup A_{n}$ be a partition of the set of positive integers. If $A_{1}$ is not good, then $A_{2} \cup A_{3} \cup ... \cup A_{n}$ is good. If $A_{2}$ is not good, then $A_{3} \cup ... \cup A_{n}$ is good etc. In the end, you get that if $A_{n-1}$ is not good, then $A_{n}$ is good.

Solution by Severius: For $ r = 2$, if some subset contains arbitrarily long blocks of consecutive numbers, we're done with $ m = 1$. Otherwise they have maximal block lengths $ M_1, M_2$, so that any block of consecutive numbers in the same subset is of length at most $ M$ for some fixed $ M$, and any two consecutive elements of a subset differ by no more than $ M + 1$. So we're done with $ m = M + 1$. Now suppose the problem is true for $ r$, and take it for $ r + 1$. Among the $ r + 1$ subsets, imagine that two of them are the same and apply the induction hypothesis. If the resulting good subset is one of the singles, we're done; otherwise, it must be the "double" subset, say, the union of $ A$ and $ B$. For each $ k$ we take a sequence $ a_1 < a_2 < ... < a_k$ contained in $ A \cup B$ and with $ - m \leq a_{i + 1} - a_i \leq m$, and we consider the collection of these sequences. If among them there are arbitrarily long subsequences of consecutive elements $ a_{i + 1}, a_{i + 2}, ..., a_{i + d}$ in the same subset, we're done. Otherwise, as above, the maximal length of such a block is bounded by some fixed $ M$, and in a sequence $ a_1 < a_2 < ... < a_k$, for any element $ a_i$, the smallest $ j > i$ such that $ a_j$ is in the same subset as $ a_i$ satisfies $ j \leq i + M + 1$, so it is easy to see that we are done.

Call a sequence of $n$ consecutive integers an $n$-string. Call a subset $A_i$ bad if it does not satisfy the property: namely, that for any $m$, there exists a $k$, such that in any $k$ sequential elements of $A_i$, at least one adjacent pair differs by more than $m$. You can verify the following consequence: for any $m$, there exists a $p$ (namely $p=mk$) such that any $p$-string contains an $m$-string with no elements from $A_i$. Now assume all $A_2$ through $A_r$ are bad. We claim $A_1$ is good: namely, we prove that for any $k$, we can find a $k$-string contained in $A_1$ (the case where $m=1$ ). Given any $k$, define the numbers $p_1,p_2, ... p_r$ as follows: $p_1=k$, and for $i>1$, $p_i$ is the number such that any $p_i$-string contains a $p_{i-1}$-string with no elements from $A_i$. The existence of these numbers is ensured by: (1) the existence of $p_1$, (2) the badness of $A_2$ through $A_r$, and (3) the consequence of badness described earlier. It follows that any $p_r$-string contains a $p_{r-1}$-string with no elements from $A_{r}$, and this string contains a $p_{r-2}$-string with no elements from $A_{r-1}$ or $A_r$, and this string contains ... etc. .... and this string contains a $p_1$-string with no elements from any of $A_2$ through $A_r$. This $p_1$-string must then be contained in $A_1$, so we are done.