1. How many irreducible fractions exist where the sum of the numerator and denominator is 100?
Answer: 20.

Solution: Consider the equation $a+b=100$. If $a$ and $b$ have a common divisor, then this divisor also divides 100, meaning that the only possible prime common divisors of $a$ and $b$ satisfying the equation can be 2 and 5. We will pair all numbers as ( $a, 100-a$ ). Since the fraction $\frac{a}{b}$ is irreducible, we exclude even numbers from 1 to 100 (leaving 50 numbers), and numbers divisible by 5 (leaving 40 numbers). Among these 40 numbers, each number $a$ corresponds to the number $100-a$, because if $a$ is odd and not divisible by 5, then $100-a$ also has these properties. We will write the smaller of the numbers $a, 100-a$ in the numerator of the fraction. This results in 20 pairs.

Comment: Correct solution - 20 points. Incomplete justification - 15 points. Correct solution with one calculation error - 18 points. Solution not completed or incorrect, but with good progress - 10 points. Solution started, with some progress - 5 points. Solution started, but with insignificant progress - 1 point. Only the correct answer without a solution - 1 point.