Xjyuk

The electric field $bfE$ represents how much force would act on a particle at a certain position per unit charge.
However, if we actually place a particle in that position, the electric field will have a singularity there (because of the $frac1r^2$ in Coulomb's law). Isn't this kind of a paradox? In my eyes, this makes the concept of electric field useless, because it cannot be used to calculate the force on a particle.

It's true that a point particle with finite charge is problematic in electromagnetism because of the infinite field and associated energy near such a particle. However, we don't need that concept in order to make a defining statement about the electric field. Rather, we can use
$$
bf E = lim_r rightarrow 0 fracbf fq
$$
where $bf f$ is the force on a charged sphere of radius $r$ with a finite charge density $rho$ independent of $r$, and $q = (4/3) pi r^3 rho$ is the charge on the sphere. This charge $q$ will tend to zero as the radius does, and it does so sufficiently quickly that no infinities arise and everything is ok.

You're forgetting one thing: a particle cannot feel its own electric field, so a point charge that generates a $1/r^2$ field doesn't do anything unless acted upon by an external field. You also can't place a particle at $r=0$ of another particle's $1/r^2$ electric field, because, well, there's already a particle there. (Also, how are you going to get it there, even if you could? It takes so much energy to even get close that you're leaving the realm of classical electromagnetism when you try.)

Isn't this kind of a paradox?

Consider two point charges, $q_1$ and $q_2$, in the vacuum with separation vector $mathbfr_12$. Coulomb's Law for the force on charge $q_2$:

$$mathbfF_2=q_2fracq_14piepsilon_0frachatmathbfr_12=q_2mathbfE_1$$

Thus, the force on charge $q_2$ is due to the electric field of charge $q_1$ only. Similarly,

$$mathbfF_1=q_1fracq_24piepsilon_0frachatmathbfr_21mathbfr_21=q_1mathbfE_2$$

the force on charge $q_1$ is due to the electric field of charge $q_2$ only. This easily generalizes to $N$ point charges; the force on charge $q_n$ is the vector sum of the forces due to electric field of each of the other $N-1$ charges.

You may (or may not) be familiar with the notion of a test charge which 'feels' the electric field due to other charges but has no significant electric field. Armed with this abstraction, one can say that the (total) electric field at a point is the force per unit charge at that point. Indeed, from the Wikipedia article Electric field

The electric field is defined mathematically as a vector field that
associates to each point in space the (electrostatic or Coulomb) force
per unit of charge exerted on an infinitesimal positive test charge at
rest at that point.

(emphasis mine)

You should distinguish the interaction if a charge with another charge from its self-interaction. For the first case there is no issue. For the second case there are issues. For a classical point particle the self-interaction energy diverges, so you will have to assume a finite radius. If you assume a homogeneus spherical distribution and equate the self energy to the rest energy you find about 2.8 femtometer for an electron. See https://en.m.wikipedia.org/wiki/Classical_electron_radius. However there is no experimental evidence for a finite value of the electron radius. As far as high energy physicists know it is a point particle.