8 If the four digits of the four-digit number $\overline{a b c d}$ satisfy $a+b=c+d$, it is called a "good number"; for example, 2011 is a "good number". Then, the number of "good numbers" is $\qquad$ .
8615 Prompt: Given $1 \leqslant a \leqslant 9, 0 \leqslant b, c, d \leqslant 9$, let $k=a+b=c+d$, then $1 \leqslant k \leqslant 18$.

When $1 \leqslant k \leqslant 9$, then in the above equation, $a$ can take any value in $\{1, \cdots, k\}$, $c$ can take any value in $\{0,1, \cdots, k\}$, and once $a, c$ are determined, $b, d$ are also determined. Therefore, the number of four-digit numbers $\overline{a b c d}$ satisfying $k=a+b=c+d$ is $k(k+1)$;
Thus, the number of four-digit numbers $\overline{a b c d}$ satisfying $k \leqslant 9$ is $\sum_{k=1}^{9} k(k+1)=330$ (numbers);
When $10 \leqslant k \leqslant 18$, from $k=a+b=c+d$ we know that $a, b, c, d$ cannot be 0. Let $a_{1}=10-a, b_{1}=10-b, c_{1}=10-c, d_{1}=10-d$, then $1 \leqslant a_{1}, b_{1}, c_{1}, d_{1} \leqslant 9$, and let $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$, then $2 \leqslant k_{1} \leqslant 10$, and the four-digit number $\overline{a b c d}$ corresponds one-to-one with the four-digit number $\overline{a_{1} b_{1} c_{1} d_{1}}$. In the above equation, $a_{1}$ and $c_{1}$ can take any value in $\left\{1, \cdots, k_{1}-1\right\}$, and once $a_{1}, c_{1}$ are determined, $b_{1}, d_{1}$ are also determined. Therefore, the number of four-digit numbers $\overline{a_{1} b_{1} c_{1} d_{1}}$ satisfying $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$ is $\left(k_{1}-1\right)^{2}$, thus the number of $\overline{a_{1} b_{1} c_{1} d_{1}}$ satisfying $2 \leqslant k_{1} \leqslant 10$ is $\sum_{k_{1}=2}^{10}\left(k_{1}-1\right)^{2}=\sum_{k=1}^{9} k^{2}$ (numbers), i.e., the number of four-digit numbers $\overline{a b c d}$ satisfying $10 \leqslant k \leqslant 18$ is $\sum_{k=1}^{9} k^{2}=285$ (numbers). Therefore, the number of "good numbers" is $330+285=615$.