Shapovalov A.V.

Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written?
Lemma. The sum of any forty consecutive numbers is not less than 80.

Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}, b_{2}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a16cba6403f5db62d5b1g-36.jpg?height=58&width=2033&top_left_y=1343&top_left_x=1)
be the remainder when divided by 40. Then the sum $a_{i+1} + a_{i+2} + \ldots + a_j$ is divisible by 40, and therefore not less than 80.

Estimate. Suppose $n > 1019$ numbers are written. By the lemma, the sum of the first $1000 = 25 \cdot 40$ of them is not less than $25 \cdot 80 = 2000$. The sum of the remaining numbers (at least 20) is not less than 20. Therefore, the total sum is not less than 2020. Contradiction.

Example. 25 groups of 1, $\ldots, 1, 41$ (in each group 39 ones and the number 41) and then 19 ones.

## Answer

1019 numbers.