1. Find all pairs of natural numbers $a, b$ for which

$$
a+\frac{66}{a}=b+\frac{66}{b} .
$$
1. By canceling the right side, we modify the given equation to the form

$$
a-b+66\left(\frac{1}{a}-\frac{1}{b}\right)=(a-b)\left(1-\frac{66}{a b}\right)=\frac{1}{a b}(a-b)(a b-66)=0 .
$$

From this, it follows that the sought pairs $(a, b)$ of natural numbers are precisely those for which $a=b$ or $a b=66$.

Thus, infinitely many pairs of natural numbers of the form $(a, b)=(k, k)$, where $k$ is any natural number, satisfy the problem. Since the number $66=2 \cdot 3 \cdot 11$ has eight divisors, there are also eight pairs $(a, b) \in\{(1,66),(2,33),(3,22),(6,11),(11,6),(22,3),(33,2),(66,1)\}$.

For a complete solution, award 6 points. For merely stating that pairs $a=b$ satisfy the problem (without mentioning the other 8 solutions), award 1 point. If the student transforms the given equation into a product form but fails to mention the additional 8 specific solutions, award 3 points. If the student omits 1, or 2 (or 3) symmetric pairs from the list of specific solutions, deduct 1, or 2 points, respectively.