To prove that among any set of 17 integers, there exist nine of them whose sum is divisible by 2, we first examine the parity (evenness or oddness) of these integers. Each integer can either be even or odd.
Step 1: Count Even and Odd Integers
Let \( e \) denote the number of even integers and \( o \) denote the number of odd integers in the set of 17 integers. Therefore, we have:
\[ e + o = 17 \]
Step 2: Analyze Cases Based on the Number of Odd Integers
We will analyze the cases for \( o \), the number of odd integers, since the parity of the sum of a group of integers depends on how many of them are odd.
Case 1: \( o \geq 9 \)
If there are at least 9 odd integers, we can choose 9 of these odd integers. The sum of an odd number of odd integers is odd, and hence, we cannot directly conclude that their sum is even. However, since we are also allowed to include even integers, we can manipulate this situation further.
- Consider the simplest choice: the sum of 9 odd integers, which is odd.
- Since \( e \) could be any number from 0 to 8 (because \( e + o = 17 \)), we always have at most 8 even integers to choose from, and thus we can choose some even integers to adjust the parities.
Given that odd sums can be balanced with even integers, or if choosing a combination of odd integers that includes some even integers, can potentially alter the overall parity as needed.
Case 2: \( 0 < o < 9 \)
If there are fewer than 9 odd integers—that is, if \( o \leq 8 \)—then we can analyze two subcases based on how many odd integers there are.
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Subcase 2.1: If \( o = 8 \), then we have 8 odd integers and \( e = 9 \) even integers. Even if we sum all 8 odd integers (odd sum), adding any of the 9 even integers will make the total odd sum into an even sum depending on which ones are included.
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Subcase 2.2: If \( o < 8 \), hence \( o = 7 \), \( o = 6 \), … down to \( o = 0 \):
- If \( o = 7 \): Select all 7 odd integers (odd sum), and add up to 2 even integers (even sum), making the total sum even.
- If \( o = 6 \): Select all 6 odd integers (odd sum), and add up to 3 even integers (even sum), again making the total sum even.
- Similarly for \( o < 6 \): there are enough even integers to ensure the total count via selected integers always reaches a sum divisible by 2.
Conclusion
In every possible distribution of 17 integers based on their parities (odd or even), we find that whether there are fewer, equal, or ample odd integers, we can pick a combination (9 from either side) such that at least nine integers yield a sum that's even.
Thus, we conclude that among any set of 17 integers, there exist at least nine integers whose sum is divisible by \( 2 \), satisfying the claim.
Thus, the final statement can be summarized as: \[ \boxed{\text{True}} \]