1. Can we find four different natural numbers, each of which is not divisible by 2, 3, or 4, but the sum of any two is divisible by 2, the sum of any three is divisible by 3, and the sum of all four is divisible by 4?
Answer: For example, 5, 17, 29, and 41.

Solution. The four numbers mentioned can be written in the form $12k + 5$, where $k$ takes the values 0, $1, 2, 3$. Therefore, the sum of any three numbers

$$
\left(12k_{1} + 5\right) + \left(12k_{2} + 5\right) + \left(12k_{3} + 5\right) = 12\left(k_{1} + k_{2} + k_{3}\right) + 15
$$

is divisible by 3. All numbers in the set are odd, so the sum of any two is divisible by 2. Finally, the sum of all four numbers is 72 and is divisible by 4.

Remark. It can be proven that all numbers have the same remainder when divided by 6.

Criteria. Any correct example - 7 points. Noted that the numbers have the same remainder when divided by 3 or 6, but an incorrect example is provided -2 points.