Is it possible to have a non-conservative vector field, such that the closed loop integral is $0$ for only some specific path(s)? I was wondering whether there exists some non-conservative field in which the closed loop integral over some specific path(s) is $0$, even if it's not $0$ for all the closed loop integrals. Or to put it in another way, is it true that any closed loop integral in a non conservative field is always non zero. 
I'd be much obliged if someone could give me an example of such a field, or at least give me some pointers.
 A: 
I'd be much obliged if someone could give me an example of such a
  field

Consider a vector field $\vec F$ with non-zero curl in the $z$ direction only:
$$\nabla \times \vec F = \left(\frac{\partial F_y}{\partial x} -  \frac{\partial F_x}{\partial y}\right)\hat z $$
An example of such a field is
$$\vec F = -y \hat x + x \hat y$$
which has curl
$$\nabla \times \vec F = 2\hat z$$
so $\vec F$ is non-conservative.  By Stoke's theorem, for a closed contour in the $z=z_0$ plane enclosing an area $A$, the line integral along that contour will have the value of $2A$.
Clearly, for plane areas parallel to the $z$ axis, there is zero flux of $\nabla \times \vec F$ through, thus, the closed contour integral of $\vec F$ in such a plane will be zero by Stoke's theorem.
To verify, pick a contour within the plane defined by, e.g., $y = 1$.
Integrate the field along the path from $(0,1,0)$ to $(1,1,0)$ to $(1,1,1)$ to $(0,1,1)$ to $(0,1,0)$
There is no component of $\vec F$ in the $z$ direction so the only contributions are the integrals along the $x$ direction.
$$\int_0^1 (-1)\hat x \cdot \hat x dx + \int_1^0 (-1)\hat x \cdot \hat x dx = -1 + 1 = 0$$
So, this is a simple example of a non-conservative vector field and a closed contour integral in that field that is zero.
A: How picky are you about singularities?
Consider the magnetic field of a steady current oriented in the $+z$ direction. The field is 
$$\vec B \sim \frac{1}{r}\hat \phi,$$
If you do a naive calculation of the curl, you'll find it's zero. And indeed, if you evaluate a closed line integral that does not bound $(x,y)=(0,0)$, you'll get zero. (Think of Ampere's law; no current enclosed) But this mistake in calculating the curl comes about from not correctly considering the $\delta$-function current source.
If your path does bound $(x,y)=(0,0)$, you'll get something that's non-zero for the closed line integral.
A: The answer is trivially positive. Take a conservative vector field $X$ in $\mathbb R^n$ and consider a closed smooth curve $\gamma$ in a open bounded region, say $\Omega\subset \mathbb R^n$. Obviously $\int_\gamma X \cdot d x =0$. Next consider another vector field $Y$ in $\mathbb R^n$ which is non-conservative  in a similar region $\Omega'$ disconnected  from $\Omega$. $\Omega'$ must contain a curve $\gamma'$ such that $\int_{\gamma'} Y\cdot d x \neq 0$, othervise $Y$ would be conservative therein. 
Finally consider a couple of smooth real valued functions $f$ and $g$ with
$f(x)=1$ in $\Omega$ and $f(x)=0$ in $\Omega'$,  and $g(x)=1$ in $\Omega'$ and $g(x)=0$ in $\Omega$ and $f(x)+g(x)=1$ everywhere (it is a trivial partition of unity, which always exists)-
The vector field $Z = fX + gY$ is smooth on the whole space $\mathbb R^n$ and it is not conservative because
$$\int_{\gamma'} Z \cdot d x  = \int_{\gamma'} Y \cdot d x \neq 0$$
however it verifies
$$\int_{\gamma} Z \cdot d x  = \int_{\gamma} X \cdot d x = 0\:.$$
