Please clarify "to the left of $b$". Does it necessary that $a$ must be adjacent to $b$?

Sorry, yes, it is necessary.

The answer is 100. Consider the general case $1,2,\ldots,2n$. If $\left|Q\right|>2n$, there are more than $\frac{(2n)!}{(2n-2)!}$ possible ways of selecting ordered number pair $(a,b)$. According to Pigeonhole Principle, at least one pair $(a_0,b_0)$ is repeated, a contradiction. When $\left|Q\right|=2n$, here is the construction (written in permutation and in modular meaning): Permutation $k(1\leq k\leq 2n)$ is \[\{k, k+1, k-1, k+2, k-2, \ldots, k+n-1, k-(n-1), k+n\}\bmod 2n\]Notice that the absolute value of difference in the neighbouring term is $1, 2, 3, \ldots, n-1, n, n-1, \ldots, 2, 1$, and we are done. Let $2n=100$ and we get what we want.

Uhh, how do we cover a complete graph \(K_{2n}\) using \(n\) paths and with each path of length \(2n-1\)?

I have a consruction for $p-1$, where $p$ is a prime number. For each $1\leq i\leq p-1$, let the $i$th permutation be $\{ik\pmod p\}_{k=1}^{p-1}$