19. (24th All-Soviet Union Mathematical Olympiad, 1990) Find the maximum value of the following expression:
$|\cdots|\left|x_{1}-x_{2}\right|-x_{3}\left|-\cdots-x_{1990}\right|$, where $x_{1}, x_{2}, \cdots, x_{1990}$ are distinct natural numbers from 1 to 1990.
19. When $x \geqslant 0, y \geqslant 0$, it is clear that $|x-y| \leqslant \max \{x, y\}$ and $\max \{\max \{x, y\}\}, z=\max \{x, y, z\}$. Therefore, when $n=1,2, \cdots, 1990$, we have
$|\cdots|\left|x_{1}-x_{2}\right|-x_{3}\left|-\cdots-x_{n}\right| \leqslant \max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, so the expression given in the problem does not exceed
$$
\max \left\{x_{1}, \cdots, x_{1990}\right\}=1990.
$$

However, this expression cannot equal 1990, because the parity of the value of this expression should be the same as the parity of the sum
$$
x_{1}+x_{2}+\cdots+x_{1990}=1+2+\cdots+1990=\frac{1990 \cdot 1991}{2}=995 \cdot 1991,
$$

which is odd. The following specific example shows that the value of the expression can reach 1989: $|\cdots||| 2-4|-5|-3 \mid-\cdots-(4 k+2)|-(4 k+4)|-|(4 k+5)|-(4 k+3)|-\cdots-1986|-1988|-1989|-1987|-1990|-1 \mid=$ 1989.
In summary, the maximum value sought is 1989.