Potential difference with an inductor As far as I know, the potential difference between two points is defined as the negative line integral of electric field between those 2 points:
$$\Delta V=-\int d \ell\cdot\mathbf E$$
I also know that when magnetic field changes, the curl of electric field is not 0 and potential difference makes no sense.
But when we have an inductor in a RLC circuit, then people always say that there is a potential drop across the ends of inductor. But since the magnetic field is changing potential difference, this should make no sense.
 A: It is not true that "potential difference makes no sense when the curl of $\vec E$ is nonzero". Instead, the scalar and vector potentials $\Phi,\vec A$ may always be defined and the electric field may be expressed in terms of these variables as
$$ E = -\nabla \phi - \frac{\partial A}{\partial t} $$
In fact, these potentials aren't unique – any other choice of the potentials related to the original one by a gauge transformation is equally good to calculate the same $\vec E$ and $\vec B = \text{curl }\vec A$.
In an rlc circuit, one may always confine the nonzero $\vec A$ to a small vicinity of the inductor which also determines the right gauge transformation. Then the potential difference across any element of the circuit is well-defined and the sum of these differences over a closed loop is zero.
A: While i think Lubos answer leaves nothing conceptualy to add, it is on too high level, if you have problem to understand nonconservative fields. Your question is actually freshman topic so i think at first you should understand this in freshman fashion and then move to Lubos answer.
Therefore you might use a very nice resource from OCW MIT by Walter Lewin
It basically tells you to always use Faraday's law. In this framework, as you said, when we have not-changing magnetic field, we get that electric potential drop on any loop is 0 (which textbooks call Kirchhoff II law). 
When we have changing magnetic field, than going in the direction of the current (so that you won't have to think about sign in inductance part), you write that the right-hand-side of Faraday's law (that is negative derivative of magnetic flux with respect to time) is simply =$-L\,dI/dt$
Than you do left-hand-side, remembering that ideal conductor has no resistance - therefore there is no electric field in it.
If you still don't get it, here is another resource, or the full lecture
