2. For each natural number $n>2$, find at least one pair of different natural numbers $p, q$ such that the number $\frac{1}{n}$ is the arithmetic mean of the numbers $\frac{1}{p}$ and $\frac{1}{q}$.
2. For any arbitrarily chosen natural number $n>2$, we seek an example of different natural numbers $p, q$ dependent on the number $n$ such that

$$
\frac{1}{n}=\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}\right)
$$

After rearrangements, this equation takes the form

$$
2 p q=n(p+q), \quad \text { or } \quad p(2 q-n)=n q
$$

Since it is sufficient to find just one pair of numbers $p, q$, it is possible to search for them by testing a few simple possibilities in the last equation.

Let's try setting $2 q-n=1$. This gives us $q=\frac{1}{2}(n+1)$ and $p=\frac{1}{2} n(n+1)$. These numbers are natural and distinct for any odd number $n>2$.

Next, let's try setting $2 q-n=2$. This gives us $q=\frac{1}{2}(n+2)$ and $p=\frac{1}{4} n(n+2)$. These numbers are natural and distinct for any even number $n>2$.

Therefore, for an odd number $n>2$, we can set $q=\frac{1}{2}(n+1)$ and $p=\frac{1}{2} n(n+1)$, and for an even number $n>2$, we can set $q=\frac{1}{2}(n+2)$ and $p=\frac{1}{4} n(n+2)$.

For a complete solution, award 6 points, of which 2 points are awarded for deriving the equation $p(2 q-n)=n q$ or another suitable equation that serves to construct the numbers $p$ and $q$.