How many numbers#1111?

The number must be a multiple of 9999.If the number is in the form of $\overline{abcdefgh}$ ,then $\overline{abcd}+\overline{efgh}=9999$ ,we get $a+e=b+f=c+g=d+h=9$ We know $1+8=2+7=3+6=4+5=9$ ,so we'll have to find the possibilities of $\overline{abcd}$ .

$a$ has 8 possibilities, $b$ has 6, $c$ has 4, $d$ has 2.So it's $8\times6\times4\times2=384$