1. Find all even natural numbers $n$ for which

$$
-53<\frac{2009}{53-n}<53-n
$$
II/1. If $n0>-53$. The inequality $\frac{2009}{53-n}<44$, implies $53-n>44$ or $9>n$. Possible values of $n$ are $2, 4, 6$, and $8$, and for all of them, $53-n \geq 45>\sqrt{2009}$, so the original inequality is satisfied.

Let $n>53$. The inequality is equivalent to

$$
-53(53-n)>2009>(53-n)^{2}
$$

From $-53^{2}+53 n>2009$ it follows that $n>\frac{2009+53^{2}}{53}=90+\frac{48}{53}$. On the other hand, $45>\sqrt{2009}>n-53$, so $n0$) in the case when $n\sqrt{2009}>53-n$ (or $n^{2}-106 n+80090+\frac{48}{53}$ in the case when $n>53$ 1 point

Checking the obtained possible values of $n$ that are greater than 53 1 point

Conclusion that the inequality is solved by $n=92, n=94$, and $n=96$ 1 point (If the contestant states that the inequality is also solved by some other even natural number $n$, 1 point is deducted.)