Let's determine all natural numbers $a, b, c, d$ for which $a b=c+d$, and $a+b=c d$ hold.
By multiplying the corresponding sides of the two equations, we get $(a+b)(c+d)=a b c d$, from which

$$
\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{c}+\frac{1}{d}\right)=1
$$

follows. If each of the numbers $a, b, c, d$ is at least 2, then

$$
\frac{1}{a}+\frac{1}{b} \leq \frac{1}{2}+\frac{1}{2}=1, \quad \text { and } \quad \frac{1}{c}+\frac{1}{d} \leq 1
$$

In this case, (1) can only be satisfied if both factors of the product on the left side are equal to 1, that is, $a=b=c=d=2$. These values also meet the requirements of the problem.

If there is a number among $a, b, c, d$ that is less than 2, let this number be $d$: $d=1$. Then, according to the original conditions, $c=a+b$, and $a b=c+1$, that is,

$$
a b=a+b+1
$$

from which

$$
(a-1)(b-1)=2
$$

follows. The number 2 can only be factored into non-negative integers as $1 \cdot 2$ or $2 \cdot 1$, so $a=2, b=3$, or $a=3, b=2$; in both cases, the value of $c$ is 5. We get the same values for $a$ and $b$ if $c=1$, but then $d=5$. If the value of $a$ or $b$ is 1, then the other is 5, and either $c=2, d=3$, or $c=3, d=2$. In total, 9 different number quadruples satisfy the conditions of the problem.