Problem 4. On the board, there are $N$ natural numbers, where $N \geqslant 5$. It is known that the sum of all the numbers is 80, and the sum of any five of them is no more than 19. What is the smallest value that $N$ can take?
Solution. Answer: 26.

The condition of the problem is equivalent to the sum of the five largest numbers not exceeding 19, and the sum of all numbers being 80. Note that among the five largest numbers, there must be a number not greater than 3 (otherwise, if all of them are not less than 4, their sum is not less than $4 \cdot 5=20$, but it should not exceed 19).

Let the total number of numbers be $x$. Since among the five largest numbers there is a number not greater than 3, then all the other $x-5$ numbers are not greater than 3. Therefore, the sum of all numbers does not exceed $3(x-5)+19$, and on the other hand, it is equal to 80. Thus, we get the inequality $3(x-5)+19 \geqslant 80$. Solving this inequality and considering that $x$ is an integer, we get that $x \geqslant 26$.

It remains to provide an example of such 26 numbers. Let's take 24 threes and 2 fours, then the total sum is $24 \cdot 3 + 2 \cdot 4 = 80$, and the sum of any five is not greater than $3 \cdot 3 + 2 \cdot 4 = 17 \leqslant 19$.

## Criteria

The most appropriate criterion is used:

7 points. Any complete solution of the problem.

3 points. 26 numbers satisfying the condition of the problem are explicitly provided.