Example 3 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad)
Proof: First, we prove that it is impossible for all 98 sums to be odd. We use proof by contradiction.

If all the sums are odd, then the arrangement of these 100 natural numbers can only be one of the following four cases:
(1) odd odd odd odd odd odd ...; (2) odd even even odd even even ...;
(3) even odd even even odd even ...; (4) even even odd even even odd ....
(1) indicates that the first 100 natural numbers are all odd, (2) and (4) indicate that the number of odd numbers among the first 100 natural numbers is more than the number of even numbers, which contradicts the fact.
Therefore, it is impossible for all 98 sums to be odd.
Next, we show that 97 of the 98 sums can be odd. We arrange the numbers from 1 to 100 in the following order:

Thus, only the 75th sum is obtained by adding even odd odd, which should be even, while the other 97 sums are all odd.
Therefore, at most 97 of the sums can be odd.