Example 4 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)
Solution: First, it is impossible for all 98 sums to be odd; otherwise, the arrangement of these 100 natural numbers could only be one of the following:
(1) 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
These four cases contradict the sequence of 100 consecutive natural numbers.
By arranging the natural numbers from $1 \sim 100$ in the following order:
odd even even odd even even……odd even even odd odd……odd
25 odd numbers 50 even numbers
25 odd numbers
We can obtain 97 odd sums. Therefore, the maximum number of sums that are odd is $97$.