There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$, $m$ divides $(a_{n+m} - a_{n})$ for all integers $n$ with $1 \leq n < n+m \leq 30$. Find $N$.
The requirement is equivalent to (where $m=2,3$, and $5$ respectively)
- All the elements with interval $m$ are belong to the same residue class of $m$
- Every block of $m$ forms a complete residue system of $m$
So for example, if we choose the order within these three residue systems as $\{0, 1\}$, $\{0, 1, 2\}$ and $\{0, 1, 2, 3, 4\}$. Then every number MOD $2$, $3$, and $5$ in this sequence will be
- $0$, $1$, $0$, $1$, $\cdots$, $0$, $1$
- $0$, $1$, $2$, $0$, $\cdots$, $1$, $2$
- $0$, $1$, $2$, $3$, $\cdots$, $3$, $4$
Therefore, element $a_n$ is determined by solving the following MOD equation $$\left\{ \begin{array}{l} a_n\equiv k_{2(n)} \pmod{2} \\ a_n\equiv k_{3(n)} \pmod{3} \\ a_n\equiv k_{5(n)} \pmod{5}\end{array} \right.$$
where $k_{2(n)}$, $k_{3(n)}$, and $k_{5(n)}$ are the corresponding residues at position $n$.
Because $2\times 3\times 5=30$ and these three numbers are pair-wise co-prime, therefore the combination of $(k_{2(n)},k_{3(n)},k_{5(n)})$ are all distinct for $1\le n\le 30$. This means that their corresponding systems of MOD equations are all distinct. By the Chinese Remainder Theorem, each system has a unique solution between $1$ and $30$. This means that the choices of the element orders in these three residue systems are independent and each will produce a unique permutation. Hence the answer is $$2!\times 3!\times 5! =\boxed{1440}$$