Problem 6. Do there exist natural numbers $a, b, c, d$, such that

$$
\frac{a}{b}+\frac{c}{d}=1, \quad \frac{a}{d}+\frac{c}{b}=2008 ?
$$
Solution. We will illustrate one of the possible ways to find numbers with the desired properties.

If we set $x=\frac{1}{b}$ and $y=\frac{1}{d}$, we will obtain the following system of equations:

$$
\begin{gathered}
a x+c y=1 \\
c x+a y=2008
\end{gathered}
$$

With respect to $x$ and $y$, this system is linear, and by a standard method, we find

$$
x=\frac{a-2008 c}{a^{2}-c^{2}}, \quad y=\frac{2008 a-c}{a^{2}-c^{2}}
$$

We seek natural numbers $a$ and $c$ such that $\frac{1}{x}$ and $\frac{1}{y}$ are also natural numbers. We observe that if we multiply each of the numbers $a$ and $c$ by $k$, then $\frac{1}{x}$ and $\frac{1}{y}$ will also increase by $k$ times. Therefore, we will proceed as follows: first, we will choose $a$ and $c$ such that $\frac{1}{x}$ is a natural number, and then we will choose a suitable natural multiplier $k$ so that $\frac{1}{y}$ also becomes a natural number.

The denominator of $\frac{1}{x}=\frac{a^{2}-c^{2}}{a-2008 c}$ suggests choosing $a=2009$ and $c=1$; then $\frac{1}{x}=2009^{2}-1$ is a natural number. With these values of $a$ and $c$, we get $\frac{1}{y}=\frac{2009^{2}-1^{2}}{2008 \cdot 2009-1}$, and the denominator suggests multiplying $a$ and $c$ by $k=2008 \cdot 2009-1$. Then $\frac{1}{y}$, as well as $\frac{1}{x}$, will be natural numbers.

We have obtained the following solution:

$$
\begin{aligned}
a=2009(2008 \cdot 2009-1) & , \quad c=2008 \cdot 2009-1 \\
b=2008 \cdot 2010(2008 \cdot 2009-1) & , \quad d=2008 \cdot 2010
\end{aligned}
$$