9.2. If Petya gives two of his notebooks to Vasya, then Vasya will have $n$ times more notebooks than Petya, and if Vasya gives $n$ of his notebooks to Petya, then Petya will have twice as many notebooks as Vasya. Find all natural values of $n$ for which this is possible.
Answer. $n=1,2,3,8$.

Solution. Let the number of notebooks that Petya and Vasya have be $x$ and $y$ respectively. After Petya gives Vasya two notebooks, we get the equation $n(x-2)=y+2$, and after Vasya gives Petya $n$ notebooks, we get the equation $x+n=2(y-n)$. From the first equation, we express $y=n x-2 n-2$ and substitute it into the second: $x+n=2 n x-6 n-4$, from which we get $x=\frac{7 n+4}{2 n-1}=3+\frac{n+7}{2 n-1}$. The fraction $\frac{n+7}{2 n-1}>0$ must be a natural number, so $2 n-1 \leq n+7$, hence $n \leq 8$. By checking the natural numbers from 1 to 8, we find that the fraction $\frac{n+7}{2 n-1}$ takes integer values for $n=1,2,3,8$. In this case, the corresponding pairs $(x, y)$ are $(11,7),(6,6),(5,7),(4,14)$.

Grading criteria. If the corresponding system of equations is set up but not solved: 2 points. If the solution is brought to the stage of $x=\frac{7 n+4}{2 n-1}: 3$ points.

If the solution is brought to the stage of $2 n-1 \leq n+7: 4$ points. If $n=1,2,3,8$ are found, but the corresponding pairs $(x, y)$ are not indicated: 6 points. Only examples for some $n: 1$ point. Examples for all $n: 2$ points.