$g(x):\mathbb{Z}\rightarrow\mathbb{Z}$ is a function that satisfies $$g(x)+g(y)=g(x+y)-xy.$$ If $g(23)=0$, what is the sum of all possible values of $g(35)$?
1. Given the function \( g(x): \mathbb{Z} \rightarrow \mathbb{Z} \) that satisfies the functional equation:
   \[
   g(x) + g(y) = g(x+y) - xy
   \]
   and the condition \( g(23) = 0 \).

2. Define a new function \( f(x) \) such that:
   \[
   f(x) = g(x) - \frac{x^2}{2}
   \]
   This transformation is chosen to simplify the given functional equation.

3. Substitute \( g(x) = f(x) + \frac{x^2}{2} \) into the original equation:
   \[
   f(x) + \frac{x^2}{2} + f(y) + \frac{y^2}{2} = f(x+y) + \frac{(x+y)^2}{2} - xy
   \]

4. Simplify the right-hand side:
   \[
   f(x) + \frac{x^2}{2} + f(y) + \frac{y^2}{2} = f(x+y) + \frac{x^2 + 2xy + y^2}{2} - xy
   \]
   \[
   f(x) + \frac{x^2}{2} + f(y) + \frac{y^2}{2} = f(x+y) + \frac{x^2}{2} + \frac{y^2}{2} + xy - xy
   \]
   \[
   f(x) + \frac{x^2}{2} + f(y) + \frac{y^2}{2} = f(x+y) + \frac{x^2}{2} + \frac{y^2}{2}
   \]

5. Cancel out the common terms \( \frac{x^2}{2} \) and \( \frac{y^2}{2} \) from both sides:
   \[
   f(x) + f(y) = f(x+y)
   \]
   This is Cauchy's functional equation, which has the general solution \( f(x) = cx \) for some constant \( c \in \mathbb{Q} \).

6. Therefore, we have:
   \[
   g(x) = f(x) + \frac{x^2}{2} = cx + \frac{x^2}{2}
   \]

7. Using the given condition \( g(23) = 0 \):
   \[
   g(23) = c \cdot 23 + \frac{23^2}{2} = 0
   \]
   \[
   23c + \frac{529}{2} = 0
   \]
   \[
   23c = -\frac{529}{2}
   \]
   \[
   c = -\frac{529}{2 \cdot 23} = -\frac{529}{46} = -\frac{23}{2}
   \]

8. Substitute \( c = -\frac{23}{2} \) back into the expression for \( g(x) \):
   \[
   g(x) = -\frac{23}{2}x + \frac{x^2}{2}
   \]
   \[
   g(x) = \frac{x^2 - 23x}{2}
   \]

9. To find \( g(35) \):
   \[
   g(35) = \frac{35^2 - 23 \cdot 35}{2}
   \]
   \[
   g(35) = \frac{1225 - 805}{2}
   \]
   \[
   g(35) = \frac{420}{2}
   \]
   \[
   g(35) = 210
   \]

The final answer is \( \boxed{210} \).