The differential's form of Gauss' Law is $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}. $$

This suggests that at every point in space, the the electric field $\vec{E}$ is determined by the charge density $\rho$ at that point.

But if charge density $\rho$ is non-zero at a point in space, it means that there is a point charge $q$ present at that point. If we now evaluate $\vec{E}$ on this point, we should get $\vec{E}=\infty $ since now we evaluating the electric field $\vec{E}$ on a point charge $q$. In other words, considering only the contribution from this point charge $q$, by Coulomb's law we get
$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{q}{0^2}=\infty.$$

Is my interpretation of Gauss' Law wrong? How can this problem be avoided?