2. Two squirrels had the same number of pine cones and the same number of cedar cones. In total, each squirrel had fewer than 25 cones. The first squirrel gathered as many pine cones as it already had and 26 cedar cones. It ended up with more pine cones than cedar cones. The second squirrel gathered as many cedar cones as it already had and ate 4 pine cones. It ended up with more cedar cones than pine cones. How many pine and cedar cones did each squirrel have initially?
Answer: 17 pine cones, 7 cedar cones.

Solution. Let the number of pine cones be $x$ and the number of cedar cones be $y$. We can write the conditions as: $\left\{\begin{array}{l}x+y=26 \\ 2 y>x-4\end{array}\right.$

Add the second and third inequalities: $2(x+y)>x+y+22$, or $x+y>22$. But $x+y=26$, so $x+y>22$ is always true. Now, we need to solve the system: $\left\{\begin{array}{l}x+y=26 \\ 2 y>x-4\end{array}\right.$

From the first equation, we have $y=26-x$. Substitute this into the second inequality: $2(26-x)>x-4$, or $52-2x>x-4$, which simplifies to $56>3x$, or $x<\frac{56}{3}$. Since $x$ and $y$ are integers, we need to find integer solutions for $x$ and $y$.

We also have $2y>x-4$, or $2(26-x)>x-4$, which simplifies to $52-2x>x-4$, or $56>3x$, or $x<\frac{56}{3}$. This gives us the inequality $\left\{\begin{array}{l}x>\frac{49}{3} \\ x<\frac{56}{3}\end{array}\right.$

In the interval $\left(\frac{49}{3} ; \frac{56}{3}\right)$, the only integer is 17. Therefore, $x=17, y=7$.

Comment. Correct solution - 20 points. Correct solution with one calculation error - 18 points. Total number of cones found, but solution not completed - 15 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Only correct answer without solution - 1 point.