Consider the set $S$ of permutations of $1, 2, \dots, 2022$ such that for all numbers $k$ in the
permutation, the number of numbers less than $k$ that follow $k$ is even.

For example, for $n=4; S = \{[3,4,1,2]; [3,1,2,4]; [1,2,3,4]; [1,4,2,3]\}$

If $|S| = (a!)^b$ where $a, b \in \mathbb{N}$, then find the product $ab$.
1. **Understanding the Problem:**
   We need to find the number of permutations of the set \(\{1, 2, \ldots, 2022\}\) such that for each number \(k\) in the permutation, the number of numbers less than \(k\) that follow \(k\) is even.

2. **Base Case:**
   Let's start with a smaller example to understand the pattern. For \(n = 4\), the set \(S\) of permutations is given as:
   \[
   S = \{[3, 4, 1, 2], [3, 1, 2, 4], [1, 2, 3, 4], [1, 4, 2, 3]\}
   \]
   We observe that each permutation satisfies the condition that for each \(k\), the number of numbers less than \(k\) that follow \(k\) is even.

3. **Inductive Step:**
   Assume that for \(n = p\), the number of such permutations is \((\left\lceil \frac{p}{2} \right\rceil!)^2\). We need to show that for \(n = p + 1\), the number of such permutations is \((\left\lceil \frac{p+1}{2} \right\rceil!)^2\).

4. **Inserting \(p+1\):**
   For all such permutations of \(\{1, 2, \ldots, p\}\), we can insert \(p+1\) in \(\left\lceil \frac{p+1}{2} \right\rceil\) places. This is because \(1, 2, \ldots, p\) are all less than \(p+1\), and we need to maintain the condition that the number of numbers less than \(k\) that follow \(k\) is even.

5. **Counting the Permutations:**
   For \(n = 2022\), we can insert each number in \(\left\lceil \frac{2022}{2} \right\rceil = 1011\) places. Therefore, the number of such permutations is:
   \[
   1 \times 1 \times 2 \times 2 \times \cdots \times 1011 \times 1011 = (1011!)^2
   \]

6. **Finding \(a\) and \(b\):**
   Given \(|S| = (a!)^b\), we identify \(a = 1011\) and \(b = 2\).

7. **Calculating the Product \(ab\):**
   \[
   ab = 1011 \times 2 = 2022
   \]

The final answer is \(\boxed{2022}\).