1. Let a complex number be written as $z=x+iy$ where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$.

2. Given that $z_1,z_2$ satisfy $|z-3|=\operatorname{Re}(z).$ Substitute $z=x+iy$: $|x+iy-3|=x$ $\sqrt{(x-3)^2+y^2}=x.$

3. Since modulus is non-negative, we must have $x\ge 0$. Squaring both sides: $(x-3)^2+y^2=x^2.$ Expand: $x^2-6x+9+y^2=x^2$ $y^2-6x+9=0$ $6x=y^2+9$ $x=\frac{y^2+9}{6}.$

   This is a parabola.

4. The two complex numbers $z_1,z_2$ lie on this parabola, and $\arg(z_1-z_2)=\frac{\pi}{4}.$ This means the vector from $z_2$ to $z_1$ has slope $\tan\frac{\pi}{4}=1.$ Hence the line joining the two points has slope $1$.

5. Let $z_1=x_1+iy_1,\qquad z_2=x_2+iy_2$ with $x_i=\frac{y_i^2+9}{6} \quad (i=1,2).$ Since the slope of the chord is $1$, $\frac{y_1-y_2}{x_1-x_2}=1$ so $y_1-y_2=x_1-x_2.$

6. Now compute $x_1-x_2$ using the parabola equation: $x_1-x_2=\frac{y_1^2-y_2^2}{6}=\frac{(y_1-y_2)(y_1+y_2)}{6}.$ Therefore, $y_1-y_2=\frac{(y_1-y_2)(y_1+y_2)}{6}.$

7. Since $z_1$ and $z_2$ are two distinct complex numbers, $y_1-y_2\ne 0$. So divide by $(y_1-y_2)$: $1=\frac{y_1+y_2}{6}.$ Hence $y_1+y_2=6.$

8. The imaginary part of $z_1+z_2$ is $\operatorname{Im}(z_1+z_2)=y_1+y_2=6.$

Therefore, the required integer is $\boxed{6}.$