5ـ Let $x_{1}, x_{2}, \cdots, x_{1990}$ be a permutation of $1,2, \cdots, 1990$, find the maximum value of $F=|\cdots| \mid x_{1}-$ $x_{2}\left|-x_{3}\right|-\cdots\left|-x_{1990}\right|$. (24th All-Russian Mathematical Olympiad Problem)
5. $F_{2}=|1-2|=1 \leqslant 2, F_{3}=|| 1-2|-3|=2 \leqslant 3, F_{4}=|| \mid 1-$ $2|-3|-4 \mid=4 \leqslant 4$. Generally, conjecture $F_{n} \leqslant n$. We will prove this by mathematical induction. First, note that when $x, y > 0$, $|x-y| \leqslant \max \{x, y\}$. Therefore, $\left|x_{1}-x_{2}\right| \leqslant \max \left\{x_{1}, x_{2}\right\}$, $||x_{1}-x_{2}\left|-x_{3}\right| \leqslant \max \left\{\max \left\{x_{1}, x_{2}\right\}, x_{3}\right\}=\max \left\{x_{1}, x_{2}, x_{3}\right\}$. Suppose $|\cdots|\left|x_{1}-x_{2}\right|-x_{3}|-\cdots|-x_{k} \mid \leqslant \max \left\{x_{1}, x_{2}, x_{3} \cdot \cdots_{2} x_{k}\right\}^{2}$, then $|\cdots| \mid x_{1}-$ $x_{2}\left|-x_{3}\right|-\cdots\left|-x_{k+1}\right| \leqslant \max \left\{\max \left\{x_{1}, x_{2}, x_{3}, \cdots, x_{k}\right\}, x_{k+1}\right\}=\max \left\{x_{1}\right.$, $\left.x_{2}, x_{3}, \cdots, x_{k}, x_{k+1}\right\}$ (1), so $F=||\left|x_{1}-x_{2}\right|-x_{3}|-\cdots|-x_{1990} \mid \leqslant$ $\max \left\{x_{1}, x_{2}, x_{3}, \cdots, x_{1990}\right\}=1990$. From the initial values, we know that the equality in inequality (1) does not always hold. In fact, the equality does not hold when $k=1990$. Because removing the absolute value signs and changing the signs of the terms does not change the parity of the expression $F$, so $F \equiv x_{1}+x_{2}+x_{3}+\cdots+x_{1990}=1+$ $2+\cdots+1990 \equiv 1(\bmod 2) .1990 \equiv 0(\bmod 2)$, so $F \leqslant 1989$. Below, we construct $x_{1}$, $x_{2}, x_{3}, \cdots, x_{1990}$, such that $F=1989$. Our strategy is to produce as many 0s as possible in the "sum". To achieve the maximum difference, the largest number must be subtracted from the smallest, i.e., $1990-1=1989$. Thus, we hope that the other 1988 numbers will cancel each other out. Consider 4 consecutive natural numbers $n+1, n+2, n+3, n+4$, we have $||n+$ $3-(n+1)|-(n+4)|-(n+2) \mid=0$. Divide $x_{1}, x_{2}, x_{3}, \cdots, x_{1988}$ into 497 groups of 4 consecutive numbers, the $k$-th group is: $\left(x_{4 k+1}, x_{4 k+2}, x_{4 k+3}, x_{4 k+4}\right)(k=0,1$, $2, \cdots, 496)$. Let $\left(x_{4 k+1}, x_{4 k+2}, x_{4 k+3}, x_{4 k+4}\right)=(4 k+2,4 k+4,4 k+5,4 k+$ 3), then $||\left|x_{4 k+1}-x_{4 k+2}\right|-x_{4 k+3}\left|-x_{4 k+4}\right|=|||(4 k+2)-(4 k+4)|-(4 k+$ $5)|-(4 k+3)|=0$. Therefore, let $x_{1989}=1990, x_{1990}=1$, then $F=\mid 1990-$ $1 \mid=1989$