Let $n$ be a positive integer. Determine the sum of all fractions $\frac{1}{x y}$ for which the numbers $x$ and $y$ are relatively prime, not greater than $n$, and their sum is greater than $n$.
Solution. Let's examine the sum $\sum \frac{1}{x y}$ for a few values of $n$. For $n=1$, there is only one pair of numbers: $\frac{1}{1 \cdot 1}=1$. If $n=2$, there are two pairs of numbers $(1 ; 2)$ and $(2 ; 1)$, the sum is $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 1}=1$. If $n=3$, there are 4 pairs of numbers, $(1 ; 3),(3 ; 1)$, $(2 ; 3)$, and $(3 ; 2)$, the sum is

$$
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 1}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 2}=1
$$

We conjecture that for any $n$, the sum $\sum \frac{1}{x y}$ is 1. We will prove this statement by induction. Assume that the inductive hypothesis is true for $n=k$. We need to show that the statement is also true for $n=(k+1)$. The sum decreases by the sum of $\frac{1}{x y}$ fractions where $x+y=k+1$, as these still satisfy $x+y>k$ but no longer satisfy $x+y>k+1$. At the same time, the sum increases by the sum of $\frac{1}{x y}$ fractions where $x=k+1$ and $y$ are coprime, and those where $y=k+1$ and $x$ are coprime. These pairs have not appeared before, as each number was smaller than $k+1$.

It is known that if $x$ and $y$ are coprime, then $(x+y ; x)=1$ and $(x+y ; y)=1$. If $(x+y)$ and $x$ had a common divisor greater than 1, then it would also divide the difference of the two numbers, i.e., $y$, which contradicts the fact that $x$ and $y$ are coprime. With this knowledge, we will show that the sum decreases by exactly the same amount as it increases. We previously saw that for every pair of numbers $(x, y)$, two pairs, $(x+y ; x)$ and $(x+y ; y)$, can be assigned, for which

$$
\frac{1}{(x+y) x}+\frac{1}{(x+y) y}=\frac{1}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x+y} \cdot \frac{x+y}{x y}=\frac{1}{x y}
$$

We also see that for each term that increases the sum, a pair $(x ; y)$ can be assigned: if $x+y=k+1$ and $x$ are coprime, then $x$ and $y$ are also coprime. Therefore, the correspondence is one-to-one. This proves that if $n$ changes from $k$ to $(k+1)$, the sum remains 1.

Thus, the sum of the fractions in the problem is 1 for every positive integer $n$.