Pavel chose two natural numbers $a, b(a \geqq b)$ and calculated the difference of their squares. He got 2007. Which pairs of numbers could Pavel have chosen?
$$
a^{2}-b^{2}=(a-b)(a+b)=2007=3^{2} \cdot 223
$$

The possible divisors of the number 2007 are: $1 ; 3 ; 9 ; 223 ; 669 ; 2007$. We will create a table. (The numbers in the third row can be obtained, for example, using the relation $a=\frac{1}{2}((a-b)+(a+b))$, and the numbers in the fourth row can then be easily calculated.)

| $a-b$ | 1 | 3 | 9 |
| :---: | :---: | :---: | :---: |
| $a+b$ | 2007 | 669 | 223 |
| $a$ | 1004 | 336 | 116 |
| $b$ | 1003 | 333 | 107 |

Pavel could choose the pairs 1004 and 1003 or 336 and 333 or 116 and 107.

[Factorization of $a^{2}-b^{2}$ into a product - 1 point, idea to find divisors of the number 2007 - 1 point, calculation of $a, b$ - 1 point for each pair, correct answer with all solutions - 1 point.]