Hong Kong competition problems, 2008

The answer is 28800.
Define
$$\begin{align*} X &= \{2x_j : 1 \le j \le 10\} = \{2, 4, 6, \dots, 20\}, \\ Y &= \{3x_j : 1 \le j \le 10\} = \{3, 6, 9, \dots, 30\}. \end{align*}$$
After removing the absolute value signs of the given expression, we get a sum of 10 terms from $X \cup Y$ minus the sum of the remaining 10 terms from $X \cup Y$.
Therefore, we must have
$$\begin{align*} S \le (30 + 27 + 24 + 21 + 18 + 15 + 20 + 18 + 16 + 14) \\ \quad - (3 + 6 + 9 + 12 + 2 + 4 + 6 + 8 + 10 + 12) \\ \quad = 131. \end{align*}$$
In the following, we will show that equality can be attained, and hence $S = 131$.
Indeed, we shall count the number of permutations such that the sum is $S$.
In order that each of $2s$ for $s = 7, 8, 9, 10$ and $3t$ for $t = 5, 6, 7, 8, 9, 10$ is the larger term of the pair in the same absolute value sign, the large numbers $7, 8, 9, 10$ cannot be adjacent terms (where $x_{10}$ and $x_1$ are considered as adjacent). Also, $5, 6$ cannot be the term immediately after the large numbers. Similarly, the small numbers $1, 2, 3, 4$ cannot be adjacent, and $5, 6$ cannot be the term immediately before the small numbers. Conversely, whenever all these conditions are satisfied, the given expression is equal to $131$.
Note that for any pair of large numbers, there must be a small number in between (possibly together with $5$ and/or $6$), and vice versa. WLOG assume $1$ is the first term. There are $3! \times 4! = 144$ ways to arrange the small numbers and the large numbers (for example, $1, 7, 2, 8, 3, 9, 4, 10$). Afterwards, we can only place $5$ and $6$ in the 4 gaps between a small number and a large number (but not between a large number and a small number). By some basic counting, we know that there are $4 \times 5 = 20$ ways to do so (4 ways to place $5$, and then 5 ways to place $6$ since there is one more position in the same gap as $5$). As we can shift all terms cyclically in 10 ways, the final answer is
$$144 \times 20 \times 10 = 28800.$$