10. (20 points) How many nine-digit numbers $\overline{a_{1} a_{2} \cdots a_{9}}$ satisfy $a_{1} \neq 0$, all digits are distinct, and
$$
a_{1}+a_{3}+a_{5}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8} .
$$
10. Let the missing digit be $a$, and
$$
\sum_{i=\pi} a_{i}=\sum_{i=m} a_{i}=t .
$$

Then $2 t=\sum_{i=1}^{9} a_{i}=45-a$.
Thus, $a \in\{1,3,5,7,9\}$.
If $a=1$, then $t=22$, we have
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$

If $a=3$, then $t=21$, we have
$$
\begin{array}{l}
21=1+4+7+9=1+5+6+9 \\
=1+5+7+8=2+4+6+9 \\
=2+4+7+8=2+5+6+8 \\
=0+4+8+9=0+5+7+9 \\
=0+6+7+8 ;
\end{array}
$$

If $a=5$, then $t=20$, we have
$$
\begin{array}{l}
20=1+2+8+9=1+3+7+9 \\
=1+4+6+9=1+4+7+8 \\
=2+3+6+9=2+3+7+8 \\
=2+4+6+8=3+4+6+7 \\
=0+3+8+9=0+4+7+9 ;
\end{array}
$$

If $a=7$, then $t=19$, we have
$$
\begin{array}{l}
19=1+3+6+9=1+4+5+9 \\
=1+4+6+8=2+3+5+9 \\
=2+3+6+8=2+4+5+8 \\
=0+2+8+9=0+4+6+9 \\
=0+5+6+8 ;
\end{array}
$$

If $a=9$, then $t=18$, we have
$$
\begin{array}{l}
18=1+2+7+8=1+3+6+8 \\
=1+4+5+8=1+4+6+7 \\
=2+3+5+8=2+3+6+7 \\
=2+4+5+7=3+4+5+6 \\
=0+3+7+8=0+4+6+8 \\
=0+5+6+7 .
\end{array}
$$

In summary, the number of nine-digit numbers that satisfy the conditions is
$$
49 A_{4}^{4} A_{5}^{5}-36 A_{4}^{4} A_{4}^{4}=120384 \text { (numbers). }
$$