Example 8 Let $x, y, z, u \in \mathbf{Z}, 1 \leqslant x, y, z, u \leqslant 10$, find the number of ordered quadruples $(x, y, z, u)$ that satisfy the following inequality: $\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}>0$
Analyzing the original inequality can be transformed into $(x z-y u)(z-x)(u-y)>0$. At this point, it is easy to get bogged down in discussions and aimless calculations of $x, y, z, u$. However, using the concept of sets to handle this problem can lead to reasonable classification and orderly discussion, making the seemingly endless classification and calculations simple and clear.

Solution: Let $f(a, b, c, d)=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}+\frac{d-a}{d+a}$

Let $A:\{(x, y, z, u) \mid 1 \leqslant x, y, z, u \leqslant 10, f(x, y, z, u)>0\}$

$B:\{(x, y, z, u) \mid 1 \leqslant x, y, z, u \leqslant 10, f(x, y, z, u)0 \Leftrightarrow \frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}>0 \\
\Leftrightarrow \frac{x-u}{x+u}+\frac{u-z}{u+z}+\frac{z-y}{z+y}+\frac{y-x}{y+x}<0 \\
\Leftrightarrow f(x, y, z, u)<0 \Leftrightarrow(x, u, z, y) \in B
\end{array}
$$

This indicates that there is a one-to-one mapping from set $A$ to set $B$, so $\quad|A|=|B|$
Next, we calculate $|C|$.
$$
\begin{array}{l}
(x, y, z, u) \in C \Leftrightarrow \frac{x z-y u}{(x+y)(z+u)}=\frac{x z-y u}{(y+z)(u+x)} \\
\Leftrightarrow(z-x)(u-y)(x z-y u)=0
\end{array}
$$

By the principle of step-by-step counting, $\left|C_{1}\right|=1000,\left|C_{2}\right|=900$
Let $x, y \cdot z, u \in\{1,2,3, \cdots, 10\}$
By the principle of step-by-step counting, the number of ordered quadruples $(x, y, z, u)$ that satisfy $x=y, z=u, x \neq z$ is 90; the number of ordered quadruples that satisfy $x=u, z=y, x \neq z$ is 90.
Verification shows that the number of unordered quadruples $(x, y, z, u)$ with $x y=z u$ and all elements being distinct is as follows:
$$
\begin{array}{l}
1 \times 6=2 \times 3,1 \times 8=2 \times 4,1 \times 10=2 \times 5 \\
2 \times 6=3 \times 4,2 \times 9=3 \times 6,2 \times 10=4 \times 5 \\
3 \times 8=4 \times 6,3 \times 10=5 \times 6.4 \times 10=5 \times 8
\end{array}
$$

Thus, $\left|C_{3}\right|=4 \times 2 \times 9+90+90=252$
Therefore, $|C|=2152$
So, $|A|=\frac{1}{2}(10000-2152)=3924$
Hence, the number of ordered quadruples $(x, y, z, u)$ that satisfy the given inequality is 3924.