3. The remainder of the division of a natural number n by 2021 is 800 more than the remainder of the division of the number n by 2020. Find the smallest such n. (A. Gолованов)
Answer. 2020.1221 = 2466 420. Solution. Let for a natural number $m$ the condition of the problem is satisfied. Then for some $d_{1}, d_{2}, d_{3}$ and $r$ the equality $m=2020 d_{1}+r=2021 d_{2}+r+800$ holds. But in this case, the equality $m-r=2020 d_{1}=2021 d_{2}+800$ also holds, so the number $m-r$ also satisfies the condition of the problem. Therefore, the sought smallest number $n$ is divisible by 2020, and when divided by 2021 gives a remainder of 800.

Note that $2020 d=2021 d-d=2021(d-1)+(2021-d)$. From this it follows that the number $2020 d$ for $d \leq 2021$ gives a remainder of $2021-d$ when divided by 2021. As $d$ increases, this remainder decreases, becoming equal to 800 when $d=1221$. Therefore, the smallest number that is a multiple of 2020 and gives a remainder of 800 when divided by 2021 is $2020 \cdot 1221$.