Let $S=\{1,2,3,...,12\}$. How many subsets of $S$, excluding the empty set, have an even sum but not an even product?

[i]Proposed by Gabriel Wu[/i]
1. **Identify the elements of \( S \) that are relevant to the problem:**
   - The set \( S \) is given as \( S = \{1, 2, 3, \ldots, 12\} \).
   - We need subsets with an even sum but not an even product. 

2. **Determine the conditions for the subsets:**
   - For a subset to have an even sum, it must contain an even number of odd elements.
   - For a subset to have an odd product, it must contain only odd elements (since the presence of any even number would make the product even).

3. **Extract the odd elements from \( S \):**
   - The odd elements in \( S \) are \( \{1, 3, 5, 7, 9, 11\} \).

4. **Count the subsets of odd elements with an even number of elements:**
   - We need to count the subsets of \( \{1, 3, 5, 7, 9, 11\} \) that have an even number of elements.
   - The number of elements in this set is 6, so we can choose 2, 4, or 6 elements.

5. **Calculate the number of such subsets using binomial coefficients:**
   - The number of ways to choose 2 elements from 6 is \( \binom{6}{2} \).
   - The number of ways to choose 4 elements from 6 is \( \binom{6}{4} \).
   - The number of ways to choose 6 elements from 6 is \( \binom{6}{6} \).

6. **Compute the binomial coefficients:**
   \[
   \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
   \]
   \[
   \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15
   \]
   \[
   \binom{6}{6} = 1
   \]

7. **Sum the results:**
   \[
   \binom{6}{2} + \binom{6}{4} + \binom{6}{6} = 15 + 15 + 1 = 31
   \]

Conclusion:
\[
\boxed{31}
\]