1. Let $z=x+iy$, where $x,y\in\mathbb{R}$. Then $\bar z=x-iy.$

2. Given $z^2+\bar z=0.$ First compute $z^2$: $z^2=(x+iy)^2=x^2-y^2+2ixy.$

3. Substitute into the equation: $x^2-y^2+2ixy + x-iy=0.$ Equating real and imaginary parts, we get: $x^2-y^2+x=0 \quad \text{(1)}$ $2xy-y=0 \quad \text{(2)}$

4. From (2): $y(2x-1)=0.$ So either:

   • Case A: $y=0$
   • Case B: $2x-1=0 \Rightarrow x=\frac12$

5. Solve each case.

   Case A: $y=0$

   From (1): $x^2+x=0$ $x(x+1)=0$ Hence: $x=0 \quad \text{or} \quad x=-1.$ So the solutions are: $z=0,\,-1.$

   Case B: $x=\frac12$

   Substitute into (1): $\left(\frac12\right)^2-y^2+\frac12=0$ $\frac14-y^2+\frac12=0$ $\frac34-y^2=0$ $y^2=\frac34$ $y=\pm \frac{\sqrt3}{2}.$ So the solutions are: $z=\frac12+\frac{\sqrt3}{2}i,\quad \frac12-\frac{\sqrt3}{2}i.$

6. Therefore, $S=\left\{0,\,-1,\,\frac12+\frac{\sqrt3}{2}i,\,\frac12-\frac{\sqrt3}{2}i\right\}.$

7. Now compute $\operatorname{Re}(z)+\operatorname{Im}(z)$ for each solution:

   • For $z=0$: $0+0=0$
   • For $z=-1$: $-1+0=-1$
   • For $z=\frac12+\frac{\sqrt3}{2}i$: $\frac12+\frac{\sqrt3}{2}$
   • For $z=\frac12-\frac{\sqrt3}{2}i$: $\frac12-\frac{\sqrt3}{2}$

8. Sum them: $0+(-1)+\left(\frac12+\frac{\sqrt3}{2}\right)+\left(\frac12-\frac{\sqrt3}{2}\right)$ $=-1+1=0.$

Hence, $\sum_{z\in S}(\operatorname{Re}(z)+\operatorname{Im}(z))=0.$