Problem 1235

Solution. It is clear that exactly one of the numbers $a_{1}$ and $b_{20}$ is 20. Without loss of generality, we can take $a_{1}=20$. Let, for example, $a_{i+1}-a_{i} \geq 2$ for some $i \in\{1,2, \ldots, 9\}$ and let $j$ be such that $b_{j}=a_{i}+1$. The sets $A^{\prime}=\left\{a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{10}^{\prime}\right\}$ and $\quad B^{\prime}=\left\{b_{1}^{\prime}, b_{2}^{\prime}, \ldots, b_{10}^{\prime}\right\}$ where $a_{k}^{\prime}=a_{k} \quad(k \neq i), a_{i}^{\prime}=a_{i}+1$, $b_{k}^{\prime}=b_{k}(k \neq j), b_{j}^{\prime}=b_{j}-1$ are disjoint subsets of $\{1,2,3, \ldots, 20\}$ and it holds

$$\left|a_{1}^{\prime}-b_{1}^{\prime}\right|+\left|a_{2}^{\prime}-b_{2}^{\prime}\right|+\ldots+\left|a_{10}^{\prime}-b_{10}^{\prime}\right|=\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\ldots+\left|a_{10}-b_{10}\right|$$

After a finite number of such steps, we arrive at the case $a_{1}=20, a_{2}=19, \ldots, a_{10}=11$, $b_{1}=1, b_{2}=2, \ldots, b_{10}=10$ where it holds

$$\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\left|a_{3}-b_{3}\right|+\ldots+\left|a_{10}-b_{10}\right|=19+17+15+\ldots+1=100$$