Let $S = \left\{ 1, \dots, 100 \right\}$, and for every positive integer $n$ define \[ 	T_n = \left\{ (a_1, \dots, a_n) \in S^n 		\mid a_1 + \dots + a_n \equiv 0 \pmod{100} \right\}. \] Determine which $n$ have the following property: if we color any $75$ elements of $S$ red, then at least half of the $n$-tuples in $T_n$ have an even number of coordinates with red elements.

[i]Ray Li[/i]
1. **Define the set \( S \) and the set \( T_n \):**
   \[
   S = \{1, 2, \ldots, 100\}
   \]
   For every positive integer \( n \), define:
   \[
   T_n = \left\{ (a_1, a_2, \ldots, a_n) \in S^n \mid a_1 + a_2 + \cdots + a_n \equiv 0 \pmod{100} \right\}
   \]

2. **Introduce generating functions:**
   Let \( R(x) \) be the generating function for the red elements and \( B(x) \) be the generating function for the blue elements:
   \[
   R(x) = \sum_{\text{red } i} x^i, \quad B(x) = \sum_{\text{blue } j} x^j
   \]

3. **Count the number of \( n \)-tuples with \( k \) red elements:**
   The number of \( n \)-tuples with \( k \) red elements is given by:
   \[
   \frac{1}{100} \sum_{x = \omega} \binom{n}{k} R(x)^k B(x)^{n - k}
   \]
   where the sum is over all 100th roots of unity \( \omega \).

4. **Define \( P \) and \( Q \):**
   Let \( P \) denote the number of \( n \)-tuples for which \( k \) is even, and \( Q \) denote the number of \( n \)-tuples for which \( k \) is odd. We aim to find all \( n \) for which \( P \geq Q \) always holds.

5. **Express \( P - Q \) using the Binomial Theorem:**
   \[
   P - Q = \frac{1}{100} \sum_{x = \omega} \left( \sum_{\text{even } k} \binom{n}{k} R(x)^k B(x)^{n - k} - \sum_{\text{odd } k} \binom{n}{k} R(x)^k B(x)^{n - k} \right)
   \]
   This simplifies to:
   \[
   P - Q = \frac{1}{100} \sum_{x = \omega} (B(x) - R(x))^n
   \]

6. **Simplify the expression using the fact that \( R(\omega) + B(\omega) = 0 \) for all roots of unity \( \omega \):**
   \[
   \sum_{x = \omega} (B(x) - R(x))^n = (B(1) - R(1))^n + \sum_{i = 1}^{99} (B(\omega^i) - R(\omega^i))^n
   \]
   Since \( R(\omega) + B(\omega) = 0 \), we have:
   \[
   B(\omega^i) - R(\omega^i) = 2B(\omega^i)
   \]
   Thus:
   \[
   \sum_{x = \omega} (B(x) - R(x))^n = (B(1) - R(1))^n + \sum_{i = 1}^{99} (2B(\omega^i))^n
   \]

7. **Evaluate the sums:**
   \[
   (B(1) - R(1))^n = (-50)^n
   \]
   \[
   \sum_{i = 1}^{99} (2B(\omega^i))^n = 2^n \sum_{i = 1}^{99} B(\omega^i)^n
   \]
   Using the roots of unity filter, \( \sum_{i = 0}^{99} B(\omega^i)^n \) counts 100 times the number of \( n \)-tuples that are blue, which is nonnegative.

8. **Combine the results:**
   \[
   \sum_{x = \omega} (B(x) - R(x))^n = (-50)^n + 2^n \sum_{i = 1}^{99} B(\omega^i)^n
   \]
   Since \( \sum_{i = 1}^{99} B(\omega^i)^n \) is nonnegative, we have:
   \[
   (-50)^n + 2^n \sum_{i = 1}^{99} B(\omega^i)^n \geq 0
   \]
   This holds for all even \( n \).

9. **Consider odd \( n \):**
   For odd \( n \), we can choose all \( 2 \pmod{4} \) numbers to be blue, making \( T = 0 \) due to impossibility. Thus, \( (-50)^n - 50^n \) would be negative, and \( P - Q < 0 \).

10. **Conclusion:**
    Since all even \( n \) work and odd \( n \) do not, our answer is all even \( n \).

The final answer is \(\boxed{\text{even } n}\)