1. An infinite scheme is given in which natural numbers are recorded as shown in the drawing. The writing of numbers starts with the unit, and then all natural numbers are recorded in the manner shown in the drawing.

| 21 | 22 | 23 | 24 | 25 | 26 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 20 | 7 | 8 | 9 | 10 |  |
| 19 | 6 | 1 | 2 | 11 |  |
| 18 | 5 | 4 | 3 | 12 |  |
| 17 | 16 | 15 | 14 | 13 | $\checkmark$ |

The row and column in which the unit is located are marked as the first row and the first column, and each other is counted from it to the left or right, i.e., up or down. For example, the number 24 is in the second column to the right and the third row up. Where is the number 2010 located in this scheme?
Solution: From the pattern, we see that $1=1^{2}=(2 \cdot 1-1)^{2}$ is in the first row and first column, the number $9=3^{2}=(2 \cdot 2-1)^{2}$ is in the second row above and the second column to the right, the number $25=5^{2}=(2 \cdot 3-1)^{2}$ is

| 21 | 22 | 23 | 24 | 25 | 26 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 20 | 7 | 8 | 9 | 10 |  |
| 19 | 6 | 1 | 2 | 11 |  |
| 18 | 5 | 4 | 3 | 12 |  |
| 17 | 16 | 15 | 14 | 13 |  |

in the third row above and the third column to the right, etc.

According to this, the number $(2 \cdot k-1)^{2}$ is in the $k$-th row above and the $k$-th column to the right. The closest number of this form to 2010 is the number $2025=45^{2}=(2 \cdot 23-1)^{2}$, which is in the 23rd row above and the 23rd column to the right. The smaller numbers from it, including 2010, are written to the left of it, and to reach 2010, we need to move 15 columns to the left in the same row. Therefore, 2010 is in the 23rd row above and 8th column to the right.