The 24th question: Given an integer $n \geq 2$, for a permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $\{1,2, \ldots, n\}$ and $1 \leq i \leq n$, let $x_{i}$ denote the maximum length of an increasing subsequence starting with $a_{i}$, and let $y_{i}$ denote the maximum length of a decreasing subsequence starting with $a_{i}$. Find the minimum possible value of $\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|x_{i}-y_{i}\right|$.
For any $1 \leq i \leq n-1$, if $a_{i}a_{i+1}$, then $x_{i} \leq x_{i+1}, y_{i} \geq y_{i+1}+1$. Therefore, $\left|x_{i}-y_{i}\right|$ and $\left|x_{i+1}-y_{i+1}\right|$ cannot both be zero, hence $\left|x_{i}-y_{i}\right|+\mid x_{i+1}-y_{i+1} \mid \geq 1$. Thus, we have:
$$
\sum_{\mathrm{i}=1}^{n}\left|x_{i}-y_{i}\right| \geq \sum_{i=1}^{\left[\frac{n}{2}\right]}\left(\left|x_{2 i-1}-y_{2 i-1}\right|+\left|x_{2 i}-y_{2 i}\right|\right)=\sum_{i=1}^{\left[\frac{n}{2}\right]} 1=\left[\frac{n}{2}\right]
$$

On the other hand, we can choose the permutations as follows to make the equality hold:
When $n=2 m$ is even, take $\left\{\begin{array}{c}a_{2 i}-1=m+i \\ a_{2 i}=m+1-i\end{array}\right.$, where $1 \leq i \leq m$;
When $n=2 m+1$ is odd, take $a_{1}=m+1,\left\{\begin{array}{l}a_{2 i}=m+1+i \\ a_{2 i+1}=m+1-i\end{array}\right.$, where $1 \leq i \leq m$.