Japan Mathematical Olympiad

Let $p_0=p_{2024}=0$ and let $t$ be a positive integer such that $p_t=2023$. The given condition gives $$\begin{array}{rl}& 2=4048-(2023+2023)\\ & =\left(p_1+\sum _{i=1}^{2022}|p_{i+1}-p_i|+p_{2023}\right)-\left(\sum _{i=0}^{t-1}(p_{i+1}-p_i)+\sum _{i=t}^{2023}(p_i-p_{i+1})\right)\\ & =\left(\sum _{i=0}^{t-1}|p_{i+1}-p_i|+\sum _{i=t}^{2023}|p_i-p_{i+1}|\right)-\left(\sum _{i=0}^{t-1}(p_{i+1}-p_i)+\sum _{i=t}^{2023}(p_i-p_{i+1})\right)\\ & =\sum _{i=0}^{t-1}\left(|p_{i+1}-p_i|-(p_{i+1}-p_i)\right)+\sum _{i=t}^{2023}\left(|p_i-p_{i+1}|-(p_i-p_{i+1})\right),\end{array}$$ thus $$f(i)=\{\begin{array}{ll}{p}_{i+1}-p_i& (0\le i\le t-1),\\ p_i-p_{i+1}& (t\le i\le 2023)\end{array}$$ satisfies ${\sum }_{i=0}^{2023}(|f(i)|-f(i))=2$. Since we have $$|a|-a=\{\begin{array}{ll}0& (a\ge 0),\\ 2|a|& (a<0)\end{array}$$ for any integer $a$, ${\sum }_{i=0}^{2023}(|f(i)|-f(i))=2$ if and only if there exists an integer $k$ with $0\le k\le 2023$ such that $f(k)=-1$ and $f(i)>0$ for every integer $i$ which is different from $k$ and $0\le i\le 2023$. Since $f(0)$, $f(t-1)$, $f(t)$ and $f(2023)$ are all positive, $k$ is neither $0$, $t-1$, $t$ nor $2023$.

When $k\ge t+1$, we have $0=p_0<p_1<\cdots <p_t>p_{t+1}>\cdots >p_k<p_{k+1}>p_{k+2}>\cdots >p_{2024}=0$ with $p_{k+1}=p_k+1$. Since $p_{k+1}<2023$, we have $p_k=p_{k+1}-1\le 2021$. Let $(A,B)$ be a pair of sets satisfying $A\cap B=\varnothing$, $A\cup B=\{1,2,\dots ,m-1,m+2,m+3,\dots ,2022\}$ for some integer $m$ with $1\le m\le 2021$. We fix such a pair $(A,B)$, and consider all permutations $p_1,p_2,\dots ,p_{2023}$ that satisfy the following conditions: $$\bullet \{p_1,p_2,\dots ,p_{t-1}\}=A,$$ $$\bullet p_t=2023,p_k=m,p_{k+1}=m+1,$$ $$\bullet \{p_{t+1},p_{t+2},\dots ,p_{k-1},p_{k+2},p_{k+3},\dots ,p_{2023}\}=B$$ The number of such permutations is to be determined. Note that $p_1,p_2,\dots ,p_{t-1}$ are arranged in increasing order of elements in $A$. The numbers $p_{t+1},p_{t+2},\dots ,p_{k-1}$ are arranged in decreasing order of the elements in $B$ that are greater than $m$, and the numbers $p_{k+2},p_{k+3},\dots ,p_{2023}$ are arranged in decreasing order of the elements in $B$ that are less than $m+1$. Therefore, $p_{t+1},p_{t+2},\dots ,p_{k-1},p_{k+2},p_{k+3},\dots ,p_{2023}$ are arranged in decreasing order of elements in $B$. Conversely, this permutation satisfies all the conditions, hence we conclude that there is exactly one such permutation. Since there are $2021$ choices for $m$, and $2^{2020}$ choices for $(A,B)$, there are $2021\cdot 2^{2020}$ permutations satisfying the conditions when $k\ge t+1$.

By the same reasoning, the number of permutations satisfying the conditions when $k\le t-2$ is also $2021\cdot 2^{2020}$. Thus, the answer is $2021\cdot 2^{2020}\cdot 2=2021\cdot 2^{2021}$.