15-15 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, then the total number of selection methods is
(A) 1050.
(B) 525.
(C) 1140.
(D) 190.
(High School Mathematics Competition in Fuzhou City, Fujian Province, China, 1993)
[Solution 1] Without loss of generality, let $a > b > d$, from $a + c = b + d$ we get $d > c$.
The number of ways to choose three numbers $a > b > c$ from $1, 2, \cdots, 20$ is $C_{20}^{3}$, but among them, those that do not meet the requirement $a + c = b + b$ are not allowed, which is equivalent to $a$ and $c$ having the same parity.

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 $C_{10}^{2}$.

Thus, the number of ways to choose three numbers $a > b > c$ from $1, 2, \cdots, 20$ such that $a + c \neq b + b$ is $C_{20}^{3} - 2C_{10}^{2} = 1050$.

Each such triplet determines a valid quadruplet, but each quadruplet $(a, b, c, d)$ appears twice, as $(a, b, c)$ and $(a, d, c)$ (since $d > c$) generate the same quadruplet. Therefore, there are $\frac{1}{2} \cdot 1050 = 525$ such quadruplets.
Thus, the answer is $(B)$.
[Solution 2] Let $a + b = c + d = k$, now consider:
When $k = 5$, there are 2 pairs of numbers that sum to 5, i.e., there are $C_{2}^{2}$ equations satisfying
$$
a + b = c + d = 5;
$$

When $k = 6$, there are also 2 pairs of numbers that sum to 6, i.e., there are $C_{2}^{2}$ equations satisfying
$$
a + b = c + d = 6;
$$

Similarly, when $k = 7, 8$, there are 3 pairs of numbers that sum to 7 or 8, i.e., there are $C_{3}^{2}$ equations;
When $k = 9, 10$, there are $C_{4}^{2}$ equations; when $k = 11, 12$, there are $C_{5}^{2}$ equations; $\cdots$; when $k = 19, 20$, there are $C_{9}^{2}$ equations; when $k = 21$, there are $C_{10}^{2}$ equations; when $k = 22, 23$, there are $C_{9}^{2}$ equations; when $k = 24, 25$, there are $C_{8}^{2}$ equations; $\cdots$; when $k = 36, 37$, there are $C_{2}^{2}$ equations.
In total, there are $4(C_{2}^{2} + C_{3}^{2} + \cdots + C_{9}^{2}) + C_{10}^{2} = 525$.
Thus, the answer is $(B)$.