1, 5, 2, and 3

Let a be the smallest of the four numbers. If a >= 2, then ab >= 2b > a + b = cd > c + d. But the problem requires that ab = c + d, so we’ve reached a contradiction. Therefore, a < 2, and since a must be a positive integer, a = 1. When a = 1, the equation ab = c + d becomes b = c + d, and the equation a + b = cd becomes 1 + b = cd, which is equivalent to b = cd – 1. These results can be combined to give 1 + c + d = cd, which can be rewritten as cd – d – c – 1 = 0, or (c – 1)(d – 1) = 2. This final equation implies that c = 2 and d = 3, or vice versa.So b=5. The only solution is (1,5,2,3), or its combinations.

Consider 2 quadratic equations $x^2-(a+b)x+ab=0$ , $x^2-(c+d)x+cd=0$ .

The second one equates to this: $x^2-abx+a+b=0$ owing to the property described at the problem. The value of these equations is $a, b, c, d$ repectively.

Since $a≠b≠c≠d$ , the sum of values and the multiple of values should be different each other. (If they are same, $a=c$ and $b=d$ )

So, $a+b>ab$ , $c+d<cd$ . Note: of course, $a+b<ab$ , $c+d>cd$ is correct. I selected the first case in my solution.

$a+b>ab$ , $ab-a-b<0$ , $(a-1)(b-1)<1$ .

Since $a$ and $b$ are positive integers, if they are 2 or more, the inequation has no values.

Thus, $a$ or $b$ must be $1$ . Let $a$ be $1$ . Then the property above is $1+b=cd$ , $c+d=b=cd-1$ , $cd-c-d=1$ , $(c-1)(d-1)=2$ .

Since $c$ and $d$ are positive integers, $c=3$ , $d=2$ or $c=2$ , $d=3$ . Thus, $b=2+3=5$ .

Therefore, the possible ordered pair is only $(1, 5, 2, 3)$ and its combinations, and thus the answer is zero(never exist).