Problem 686

8. 615 .

Let $k=a+b=c+d$.
Since $1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant 9$, then
$1 \leqslant k \leqslant 18$.
When $1 \leqslant k \leqslant 9$, in the above equation, $a$ can take any value in $\{1,2$, $\cdots, k\}$, $c$ 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 $\overline{a b c d}$ that satisfy the condition is $k(k+1)$.
When $10 \leqslant k \leqslant 18$, $a 、 b 、 c 、 d$ cannot be 0.
$$\begin{array}{l} \text { Let } a_{1}=10-a, b_{1}=10-b, \\ c_{1}=10-c, d_{1}=10-d . \end{array}$$

Then $1 \leqslant a_{1} 、 b_{1} 、 c_{1} 、 d_{1} \leqslant 9$.
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 $\left\{1,2, \cdots, k_{1}-1\right\}$, and once $a_{1} 、 c_{1}$ are determined, $b_{1} 、 d_{1}$ are also determined.

Therefore, the number of four-digit numbers $\overline{a_{1} b_{1} c_{1} d_{1}}$ that satisfy $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$ is $\left(k_{1}-1\right)^{2}$.
Thus, the total number of good numbers is
$$\begin{array}{l} \sum_{k=1}^{9} k(k+1)+\sum_{k_{1}=2}^{10}\left(k_{1}-1\right)^{2} \\ =330+285=615 . \end{array}$$