7.4 On the table, there are 2020 boxes, some of which contain candies, while the others are empty. On the first box, it is written: "All boxes are empty." On the second: "At least 2019 boxes are empty." On the third: "At least 2018 boxes are empty," and so on, up to the 2020th, which says: "At least one box is empty." It is known that the inscriptions on the empty boxes are false, while those on the boxes with candies are true. Determine how many boxes contain candies. Justify your answer.
Solution. Note that if the inscription "At least $A$ boxes are empty" is true for some $A$, then all subsequent inscriptions are also true. This means that the boxes with candies are all boxes starting from some number $N$. Then the number of boxes without candies is exactly $N-1$. The inscription on the $N$-th box reads: "At least $2021-N$ boxes are empty." This is true. The inscription on the previous box, numbered $N-1$, reads: "At least $2021-(N-1)$ boxes are empty." This is false. Therefore, the number of empty boxes is $2021-N$. From the equation $N-1=2021-N$, we find that $N=1011$. Thus, the number of empty boxes is 1010, and the number of boxes with candies is $2020-1010=1010$.

Answer: 1010 boxes.

| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| There is an arithmetic error in the correct solution process, possibly leading to an incorrect answer | 6 points |
| A correct example is provided (the first 2010 boxes are empty, the rest have candies), but it is not proven that there is no example with a different number of filled boxes | 3 points |
| It is proven that the numbers of empty (full) boxes are all natural numbers not exceeding (greater than) a fixed number | 2 points |
| Correct answer without justification or with incorrect justification | 0 points |