I read that in dimension $\geq 4$ there are Gorenstein abelian quotient singularities that have no crepant resolutions. What is the simplest example? I immagine that there should be toric examples. Is it the case that the cone does not admit a suitable subdivision in simplicial cones, corresponding to a crepant resolution? In your answers, please consider that I am only an amateur algebraic geometer, but I know some toric geometry.
[EDIT: added proof that $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor.] Definition A variety is called $\mathbb Q$factorial if every Weil divisor on it is $\mathbb Q$Cartier, i.e., some multiple of it is a Cartier divisor. The general statement you might want is that
Proof $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor, and being terminal implies that all the discrepancies are positive. $\square$

Addendum Here is a proof that $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor:
Proof Let $y\in E$ and suppose that $\mathrm{codim}_YE\geq 2$ in a neighborhood of $y$. Let $C\subseteq E$ be an arbitrary proper curve such that $f(C)$ is a point and $y\in C$ and let $H\subseteq Y$ be an effective divisor such that $y\in H$, but $C\not\subseteq H$. This implies that $H\cdot C>0$. Consider the Weil(!) divisor $f_*H$ on $X$ (the pushforward is meant as a cycle). As $X$ is $\mathbb Q$factorial, some multiple of $f_*H$ will be Cartier, so replacing $H$ with that multiple we may assume that actually $f_*H$ is Cartier. Then it makes sense to pull it back (as a Cartier divisor). So we get a (Cartier) divisor $f^*f_*H$ which agrees with $H$ on $Y\setminus E$. In particular, if $\mathrm{codim}_YE\geq 2$ in a neighborhood $U$ of $y$, then $H_U=(f^*f_*H)_U$. Now by construction $y\in C\cap U\neq\emptyset$, so along $C$, $f^*f_*H=H+F$ where $F$ is an effective (exceptional) divisor that does not contain $C$. Finally, this leads to a contradiction, because we get that $$ 0=f^*f_*H\cdot C \geq H\cdot C>0$$ since $f(C)$ is a point. 


The simplest example is $\mathbb{C}^4/\pm1$ where $1$ acts diagonally. I'm sure there is an elementary proof that this does not have a crepant resolution, maybe via toric geometry. Someone else on MO might know a reference. The proof I know uses a result of Yasuda which says that if $X$ is a Gorenstein orbifold and $Y\to X$ is a crepant resolution, then $H^*_{orb}(X) = H^*(Y)$ as graded vector spaces. In the case at hand, this would imply that the exceptional fiber of a crepant resolution $Y\to \mathbb{C}^4/\pm1$ would have cohomology in degree 0 and in degree 4 only, which is impossible for a projective variety. 


There is a nice explicit description, in this paper of Morrison and Stevens, of four dimensional cyclic quotient singularities that are Gorenstein and terminal (cf. this MO questionterminal implies the nonexistence of crepant resolutions). 

