Given the set $S$ whose elements are zero and the even integers, positive and negative.  Of the five operations applied to any pair of elements:  (1) addition (2) subtraction (3) multiplication (4) division (5) finding the arithmetic mean (average), those elements that only yield elements of $S$ are:
$ \textbf{(A)}\ \text{all} \qquad\textbf{(B)}\ 1,2,3,4\qquad\textbf{(C)}\ 1,2,3,5\qquad\textbf{(D)}\ 1,2,3\qquad\textbf{(E)}\ 1,3,5 $
1. **Addition**: 
   - Consider two elements from the set \( S \), say \( 2n \) and \( 2k \), where \( n, k \in \mathbb{Z} \).
   - The sum of these two elements is \( 2n + 2k = 2(n + k) \).
   - Since \( n + k \) is an integer, \( 2(n + k) \) is an even integer, which is in the set \( S \).
   - Therefore, addition always yields an element of \( S \).

2. **Subtraction**:
   - Consider two elements from the set \( S \), say \( 2n \) and \( 2k \), where \( n, k \in \mathbb{Z} \).
   - The difference of these two elements is \( 2n - 2k = 2(n - k) \).
   - Since \( n - k \) is an integer, \( 2(n - k) \) is an even integer, which is in the set \( S \).
   - Therefore, subtraction always yields an element of \( S \).

3. **Multiplication**:
   - Consider two elements from the set \( S \), say \( 2n \) and \( 2k \), where \( n, k \in \mathbb{Z} \).
   - The product of these two elements is \( 2n \cdot 2k = 4nk \).
   - Since \( 4nk \) is an even integer, it is in the set \( S \).
   - Therefore, multiplication always yields an element of \( S \).

4. **Division**:
   - Consider two elements from the set \( S \), say \( 2n \) and \( 2k \), where \( n, k \in \mathbb{Z} \).
   - The division of these two elements is \( \frac{2n}{2k} = \frac{n}{k} \).
   - For \( \frac{n}{k} \) to be an integer, \( k \) must divide \( n \) exactly. However, if \( k = 0 \), the division is undefined.
   - Therefore, division does not always yield an element of \( S \).

5. **Arithmetic Mean**:
   - Consider two elements from the set \( S \), say \( 2n \) and \( 2k \), where \( n, k \in \mathbb{Z} \).
   - The arithmetic mean of these two elements is \( \frac{2n + 2k}{2} = n + k \).
   - For \( n + k \) to be in \( S \), it must be an even integer. However, \( n + k \) can be odd if one of \( n \) or \( k \) is odd and the other is even.
   - Therefore, the arithmetic mean does not always yield an element of \( S \).

Conclusion:
- The operations that always yield elements of \( S \) are addition, subtraction, and multiplication.

The final answer is \( \boxed{\textbf{(D)}} \)