Proof:

Let's assume that a and b are two odd positive integers, where a > b.

We need to prove that one of the numbers a - b/2 and a + b/2 is odd, while the other is even.

Case 1: a is odd and b is odd
In this case, a can be expressed as a = 2k + 1, where k is a positive integer.
Similarly, b can be expressed as b = 2m + 1, where m is a positive integer.

Now let's consider the number a - b/2:
a - b/2 = (2k + 1) - (2m + 1)/2
= (2k + 1) - (2m + 1)/2
= 2k + 1 - m - 1/2
= 2k - m/2
= 2(k - m/2)

Since k - m/2 is an integer, we can rewrite a - b/2 as 2 multiplied by an integer, which means it is even.

Similarly, let's consider the number a + b/2:
a + b/2 = (2k + 1) + (2m + 1)/2
= 2k + 1 + m + 1/2
= 2k + m/2
= 2(k + m/2)

Again, since k + m/2 is an integer, we can rewrite a + b/2 as 2 multiplied by an integer, which means it is even.

Therefore, in this case, one of the numbers a - b/2 and a + b/2 is even, while the other is odd.

Case 2: a is odd and b is even
In this case, a can be expressed as a = 2k + 1, where k is a positive integer.
Similarly, b can be expressed as b = 2m, where m is a positive integer.

Now let's consider the number a - b/2:
a - b/2 = (2k + 1) - (2m)/2
= (2k + 1) - m
= 2k + 1 - m
= 2k + (1 - m)

Since k and (1 - m) are integers, we can rewrite a - b/2 as 2 multiplied by an integer, which means it is even.

Similarly, let's consider the number a + b/2:
a + b/2 = (2k + 1) + (2m)/2
= 2k + 1 + m
= 2k + (1 + m)

Since k and (1 + m) are integers, we can rewrite a + b/2 as 2 multiplied by an integer, which means it is even.

Therefore, in this case as well, one of the numbers a - b/2 and a + b/2 is even, while the other is odd.

Conclusion:
In both cases, we have shown that one of the numbers a - b/2 and a + b/2 is even, while the other is odd. This holds true