Shapovalov A.V.

For any natural $n>2009$, from the fractions $\frac{1}{n}, \frac{2}{n-1}, \frac{3}{n-2}, \ldots, \frac{n-1}{2}, \frac{n}{1}$, can two pairs of fractions with the same sums be chosen?
Each of the given fractions has the form $\frac{n+1-a}{a}=\frac{n+1}{a}-1$, where $1 \leq a \leq n$. Therefore, it is sufficient to find such distinct natural numbers $a, b, c$ and $d$, not greater than 2009, for which

$\left(\frac{n+1}{a}-1\right)+\left(\frac{n+1}{b}-1\right)=\left(\frac{n+1}{c}-1\right)+\left(\frac{n+1}{d}-1\right)$. Removing the minus ones and then dividing by $n+1$,

we obtain the equivalent equation $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$. It remains to find fractions that satisfy this equation. This can be done by taking any equality of two sums of distinct natural addends, the LCM of which is not greater than 2009, and dividing it by this LCM. For example, dividing the equality $1+4=2+3$ by 12, we get $\frac{1}{12}+\frac{1}{3}=\frac{1}{6}+\frac{1}{4}$.