The statement that the mean is greater than the median is simply a statement of 
$$F_{X}(\mathbb{E}[X])>0.5$$
for a random variable $X$ with distribution function $F_{X}(x)$.

This has nothing (directly) to do with skewness.

To impose a direct relationship between this statement and skewness (being positive) would just result in some inequality like 

$$\mathbb{E}\bigg[\bigg(\frac{X-\mu}{\sigma}\bigg)^{3}\bigg]>0\quad\Rightarrow\quad \mathbb{E}[X^3]>\mathbb{E}[X](3σ^2+\mathbb{E}[X]^2)$$

where $\mathbb{E}[X]>F^{-1}_X(0.5)$ and $σ^2$ is taken to be finite. 

Any distribution satisfying this inequality is what you're after.