4. 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, then the total number of selection methods is ( . .
(A) 1050
(B) 525
(C) 1140
(D) 190
4. Let's assume $a>b>d$. From $a+c=b+d$ we get $d>c$.

Selecting three numbers $a>b>c$ from $1,2, \cdots, 20$ has $C_{20}^{3}$ ways, but among them, those with $a+c=b+b$ are not allowed, which is equivalent to $a$ and $c$ having the same parity. Selecting two odd numbers from $1,2, \cdots, 20$ has $C_{10}^{1}$ ways, and selecting two even numbers has $C_{10}^{3}$ ways.

Therefore, selecting three numbers $a>b>c$ from $1,2, \cdots, 20$ such that $a+c \neq b+b$ has $C_{20}^{3}-2 C_{10}^{2}=1050$ ways.

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 $d>c$). Thus, there are $\frac{1}{2} \times 1050=525$ ways. Hence, the answer is B.