The first phase of the solution consists in showing that $kl \ge n$. For every $i = 1, \dots, n$, let $f(i)$ denote the length of the longest increasing subsequence ending with $a_i$, and let $g(i)$ be the length of the longest decreasing subsequence ending with $a_i$. For distinct indices $i < j$, if $a_i < a_j$ then $f(i) < f(j)$, and if $a_i > a_j$ then $g(i) < g(j)$. Hence pairs of the form $(f(i), g(i))$ where $i = 1, \dots, n$ are all distinct, i.e., there are $n$ different such pairs in total. By the conditions of the problem, the largest number of the form $f(i)$ is $k$ and the largest number of the form $g(i)$ is $l$. Thus the number of pairs of the form $(f(i), g(i))$ is at most $kl$. Consequently, $n \le kl$.

From this result, we deduce $k+l \ge 2\sqrt{kl} \ge 2\sqrt{n}$ by AM-GM. At most one number can belong to an increasing and a decreasing subsequence simultaneously. Thus subsequences $(a_{i_1}, \dots, a_{i_k})$ and $(a_{j_1}, \dots, a_{j_l})$ together contain at least $2\sqrt{n}-1$ different natural numbers in total. By assumptions, number $n$ has at most $\lfloor\sqrt{n}\rfloor-1$ divisors that are not larger than $\sqrt{n}$, the total number $\delta(n)$ of divisors of $n$ satisfies the inequality $\delta(n) \le 2\lfloor\sqrt{n}\rfloor - 2$. Consequently, subsequences $(a_{i_1}, \dots, a_{i_k})$ and $(a_{j_1}, \dots, a_{j_l})$ together contain at least one number that does not divide $n$.

The inequality $kl \ge n$ can be proven also in the following way. Let us partition the permutation $(a_1, \dots, a_n)$ into decreasing subsequences using the following algorithm. The first element of each new subsequence is the first unused element in the original permutation, the next is the first following to it in the original permutation unused element smaller than it etc., until no more elements can be chosen this way. Let these subsequences be $K_1, \dots, K_x$ in the order of forming.

For every $z = x, x-1, \dots, 2$ and every element $a_j$ of $K_z$, there exists an element $a_i$ in $K_{z-1}$ such that $i < j$ and $a_i < a_j$. Indeed, suppose the contrary. Then all elements $a_i$ of $K_{z-1}$ such that $i < j$ are greater than $a_j$. This means that $a_j$ should have been chosen into $K_{z-1}$, a contradiction.

Hence, starting from an arbitrary element $b_x$ of $K_x$, we can choose an element $b_{x-1}$ from $K_{x-1}$, an element $b_{x-2}$ from $K_{x-2}$, etc, until $b_1$ from $K_1$, in such a way that $b_1 < \dots < b_{x-1} < b_x$. This is an increasing subsequence of length $x$ of the original permutation. As every element of the original permutation belongs to one of $K_1, \dots, K_x$, there exists a decreasing subsequence of length at least $\frac{n}{x}$. Now $k \ge x$ and $l \ge \frac{n}{x}$ together give $kl \ge x \cdot \frac{n}{x} = n$.

Another algorithm can be used for partitioning the permutation $(a_1, \dots, a_n)$ into decreasing subsequences in such a way that there exists an increasing subsequence with each element representing a different part. Let the first element of each new subsequence be the largest among the unused elements, the next be the largest following to it in the original sequence unused element etc., until no more elements can be chosen this way. Let these subsequences be $L_1, \dots, L_y$ in the order of forming.

For every $z = y, y-1, \dots, 2$ and every $a_i$ from $L_z$, there exists an element $a_j$ from $L_{z-1}$ such that $i < j$ and $a_i < a_j$. Indeed, suppose the contrary. Then all elements $a_j$ in $L_{z-1}$ such that $i < j$ are smaller than $a_i$. This means that $a_i$ should have been chosen into $L_{z-1}$, contradiction.

Hence, starting from an arbitrary element $c_y$ of $L_y$, we can choose an element $c_{y-1}$ from $L_{y-1}$, an element $c_{y-2}$ from $L_{y-2}$ etc., until $c_1$ from $L_1$, in such a way that $c_y < c_{y-1} < \dots < c_1$. The rest is as in Solution 2.