1. Find all quadruples of natural numbers $a, b, c, d$ for which

$$
\begin{aligned}
& a b+c d=1999 \\
& a c+b d=1999 \\
& a d+b c=1999
\end{aligned}
$$
1. By subtracting the second equation from the first and the third equation from the second, we obtain two equations which, after a simple rearrangement, have the form

$$
(a-d)(b-c)=0, \quad(a-b)(c-d)=0
$$

From this, it follows that at least three of the four numbers $a, b, c, d$ are equal. Let's assume, for example, that $a=b=c$. Substituting into the first equation of the original system, we get

$$
a^{2}+a d=a(a+d)=1999 .
$$

Since 1999 is a prime number, the only solution is $a=1$ and $a+d=1999$. From this, it follows that

$$
a=b=c=1, \quad d=1998 .
$$

By permutation, we obtain three more solutions

$$
a=b=d=1, c=1998, \quad a=c=d=1, b=1998, \quad b=c=d=1, a=1998 .
$$

Substitution easily confirms that all four quadruples satisfy the equations.

For a complete solution, 6 points are awarded. If the solution is found by estimation, i.e., if the proof that no other solutions exist is missing, award 2 points. If not all solutions are provided, award 4 points.