Solution below.

Reveal solution by highlighting:

This one became very easy, after I broke down and wrote a program to brute force it. Note that a) I only display 1 solution, assuming there is only 1, b) I didn’t start from 000000000 every time, but threw away solutions, after testing they would never work, because early conditions couldn’t work.

#!/usr/local/bin/php
<?php
	// abc
	// def
	// ghi
for($a=5;$a<=9;$a++) { // testing revealed at least 5
for($b=1;$b<=9;$b++) {
for($c=1;$c<=9;$c++) {
for($d=1;$d<=9;$d++) {
for($e=0;$e<=9;$e++) {
for($f=0;$f<=9;$f++) {
for($g=1;$g<=9;$g++) {
for($h=0;$h<=9;$h++) {
for($i=0;$i<=9;$i++) {

	$a1=$a*100+$b*10+$c;
	$a4=$d*100+$e*10+$f;
	$a5=$g*100+$h*10+$i;

	$d1=$a*100+$d*10+$g;
	$d2=$b*100+$e*10+$h;
	$d3=$c*100+$f*10+$i;

	$good=1;

// Across
// 1	Today's number.	(3)
// 4	A multiple of 5 times the sum of the digits of 2D.	(3)

	if(!is_int($a4/(5*($b+$e+$h)))) {
		$good=0;
	}

// 5	Greater than two times 3D.	(3)

	if($a5<=2*$d3) {
		$good=0;
	}

// Down
// 1	The sum of 5A and 3D.	(3)

	if($d1!=$a5+$d3) {
		$good=0;
	}

// 2	Nine times the sum of the digits of 4A.	(3)

	if($d2!=9*($d+$e+$f)) {
		$good=0;
	}

// 3	Greater than 2D.	(3)

	if($d3<=$d2) {
		$good=0;
	}

	if($good==1) {
		print "$a$b$c $d$e$f $g$h$i";
		exit();
	}
} // i
} // h
} // g
} // f
} // e
} // d
} // c
} // b
} // a
?>

Solution:

8	1	2
7	6	5
6	2	8

So the solution is 812.

Interestingly, there was an early version of this puzzle. Maybe it was harder (too hard), maybe it was just some formatting error.