Example 2.2.9 Find the number of all ordered quadruples of integers $(x, y, z, u)$ that satisfy $\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}>0$, and $1 \leqslant x, y, z, u$ $\leqslant 10$.
Let $f(a, b, c, d)=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}+\frac{d-a}{d+a}$, and denote
$$
\begin{array}{l}
A=\{(x, y, z, u) ; 1 \leqslant x, y, z, u \leqslant 10, f(x, y, z, u)>0\}, \\
B=\{(x, y, z, u) ; 1 \leqslant x, y, z, u \leqslant 10, f(x, y, z, u)0, \\
\frac{x-u}{x+u}+\frac{u-z}{u+z}+\frac{z-y}{z+y}+\frac{y-x}{y+x}<0,
\end{array}
$$

i.e., $(x, u, z, y) \in B$,
Let $F(x, y, z, u)=(x, u, z, y)$, then $F$ is a one-to-one mapping from $A \rightarrow B$, hence $|A|=|B|$. Now calculate $|C|$,
$$
\begin{aligned}
(x, y, z, u) \in C & \Leftrightarrow \frac{x z-y u}{(x+y)(z+u)}=\frac{x z-y u}{(y+z)(u+x)} \\
& \Leftrightarrow(z-x)(u-y)(x z-y u)=0 .
\end{aligned}
$$

Let $C_{1}=\{(x, y, z, u) ; x=z, 1 \leqslant x, y, z, u \leqslant 10\}$,
$$
\begin{array}{l}
C_{2}=\{(x, y, z, u) ; x \neq z, y=u, 1 \leqslant x, y, z, u \leqslant 10\}, \\
C_{3}=\{(x, y, z, u) ; x \neq z, y \neq u, x z=y u, 1 \leqslant x, y, z, u \leqslant 10\} .
\end{array}
$$

Since the pairs $(a, b, c, d)$ that satisfy $a \times b=c \times d$ and are distinct unordered quadruples of $1,2,3, \cdots, 10$ are only nine:
$$
\begin{array}{l}
1 \times 6=2 \times 3,1 \times 8=2 \times 4,1 \times 10=2 \times 5, \\
2 \times 6=3 \times 4,2 \times 9=3 \times 6,2 \times 10=4 \times 5, \\
3 \times 8=4 \times 6,3 \times 10=5 \times 6,4 \times 10=5 \times 8 .
\end{array}
$$

Each unordered quadruple corresponds to 8 ordered quadruples:
$$
a \times b=c \times d, b \times a=c \times d, a \times b=d \times c, b \times a=d \times c .
$$

Including the swapping of both sides of the equation, so the total number of distinct ordered quadruples is $8 \times 9=72$, all 72 quadruples belong to $C_{3}$, but $C_{3}$ can also include quadruples where $x=y \neq z=u, x=u \neq y=$ $z$, and there are $90 \times 2$ such quadruples, so
$$
\left|C_{3}\right|=72+90 \times 2=252 .
$$

It is easy to see that $\left|C_{1}\right|=1000,\left|C_{2}\right|=900$, so $|C|=\left|C_{1}\right|+\left|C_{2}\right|+\left|C_{3}\right|=2152$, therefore $|A|=\frac{10^{4}-2152}{2}=3924$.