Example 4 (1993 Fuzhou City, Fujian Province High School Competition Question) Select four different numbers $a, b, c, d$ from $\{1,2,3,4, \cdots, 20\}$, satisfying $a+c=b+d$. If the order of $a, b, c, d$ is not considered, the total number of selection methods is ( ).
A. 1050
B. 525
C. 1140
D. 190
Choose B. Reason: Suppose $a>b>d$, from $a+c=b+d$ we get $d>c$.
The number of ways to choose 3 numbers $a>b>c$ from $1,2, \cdots, 20$ is $C_{20}^{3}$, but among these, those that satisfy $a+c=b+b$ do not meet the requirement, which is equivalent to selecting 2 numbers $a, c$ from $1,2, \cdots, 20$ that are both odd or both even.

The number of ways to choose two odd numbers from $1,2, \cdots, 20$ is $C_{10}^{2}$, and the number of ways to choose two even numbers is also $C_{10}^{2}$. Therefore, the number of ways to choose 3 numbers $a>b>c$ from $1,2, \cdots, 20$ such that $a+c \neq b+b$ is $C_{20}^{3}-2 C_{10}^{2}=1050$. At this point, $d=a+c-b$ is uniquely determined. Thus, each such triplet determines a qualifying quadruplet, but each quadruplet $(a, b, c, d)$ appears twice (as $(a, b, c)$ and $(a, d, c)$, because $d>c$), resulting in $\frac{1}{2} \cdot 1050=525$ distinct quadruplets.