The zeroes of the function $f(x)=x^2-ax+2a$ are integers.  What is the sum of all possible values of $a$?

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18$
1. Let \( p \) and \( q \) denote the roots of the equation \( x^2 - ax + 2a = 0 \). By Vieta's formulas, we have:
   \[
   p + q = a
   \]
   \[
   pq = 2a
   \]

2. Substitute \( a \) from the first equation into the second equation:
   \[
   pq = 2(p + q)
   \]

3. Rearrange the equation:
   \[
   pq - 2p - 2q = 0
   \]

4. Add 4 to both sides to factorize:
   \[
   pq - 2p - 2q + 4 = 4
   \]

5. Factorize the left-hand side:
   \[
   (p - 2)(q - 2) = 4
   \]

6. Since \( p \) and \( q \) are integers, the possible pairs \((p-2, q-2)\) that multiply to 4 are:
   \[
   (1, 4), (4, 1), (-1, -4), (-4, -1), (2, 2), (-2, -2)
   \]

7. Convert these pairs back to \((p, q)\):
   \[
   (p, q) = (3, 6), (6, 3), (1, -2), (-2, 1), (4, 4), (0, 0)
   \]

8. Calculate \( a \) for each pair:
   \[
   \begin{align*}
   (3, 6) &\implies a = 3 + 6 = 9 \\
   (6, 3) &\implies a = 6 + 3 = 9 \\
   (1, -2) &\implies a = 1 - 2 = -1 \\
   (-2, 1) &\implies a = -2 + 1 = -1 \\
   (4, 4) &\implies a = 4 + 4 = 8 \\
   (0, 0) &\implies a = 0 + 0 = 0 \\
   \end{align*}
   \]

9. Sum all distinct possible values of \( a \):
   \[
   9 + (-1) + 8 + 0 = 16
   \]

The final answer is \(\boxed{16}\).