62 From $\{1,2,3,4, \cdots, 20\}$, select four different numbers $a, b, c, d$, 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. 1140
C. 525
D. 190
62 C. Suppose $a>b>d$, because
$$
a+c=b+d,
$$

so $d>c$.
The number of ways to select three numbers $a>b>c$ from $1,2, \cdots, 20$ is $C_{20}^{3}$. However, this includes cases where:
$$
a+c=b+b.
$$

In such cases, the value of $d$ derived from $a, b, c$ is the same as $b$, which is not allowed. In these cases, $a$ and $c$ have the same parity. The number of ways to select two odd numbers or two even numbers from $1,2, \cdots, 20$ is $2 C_{10}^{2}$. Therefore, the number of ways to select three numbers $a, b, c$ such that $a>b>c$ and $a+c \neq b+b$ is
$$
C_{20}^{3}-2 C_{10}^{2}=1050 \text { (ways). }
$$

Each such triplet can determine a valid quadruplet. However, each quadruplet $(a, b, c, d)$ can be obtained from both $(a, b, c)$ and $(a, d, c)$, so the total number of quadruplets is
$$
\frac{1}{2} \times 1050=525 \text { (ways). }
$$