4. Assuming that $A_{1}$ is not such a set $A_{i}$, it follows that for every $m$ there exist $m$ consecutive numbers not in $A_{1}$. It follows that $A_{2} \cup A_{3} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of numbers. Inductively, let us assume that $A_{j} \cup A_{j+1} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of consecutive numbers and none of $A_{1}, A_{2}, \ldots, A_{j-1}$ is the desired set $A_{i}$. Let us assume that $A_{j}$ is also not $A_{i}$. Hence for each $m$ there exists $k(m)$ such that among $k(m)$ elements of $A_{j}$ there exist two consecutive elements that differ by at least $m$. Let us consider $m \cdot k(m)$ consecutive numbers in $A_{j} \cup \cdots \cup A_{r}$, which exist by the induction hypothesis. Then either $A_{j}$ contains fewer than $k(m)$ of these integers, in which case $A_{j+1} \cup \cdots \cup A_{r}$ contains $m$ consecutive integers by the pigeonhole principle or $A_{j}$ contains $k(m)$ integers among which there exists a gap of length $m$ of consecutive integers that belong to $A_{j+1} \cup \cdots \cup A_{r}$. Hence we have proven that $A_{j+1} \cup \cdots \cup A_{r}$ contains sequences of integers of arbitrary length. By induction, assuming that $A_{1}, A_{2}, \ldots, A_{r-1}$ do not satisfy the conditions to be the set $A_{i}$, it follows that $A_{r}$ contains sequences of consecutive integers of arbitrary length and hence satisfies the conditions necessary for it to be the set $A_{i}$.