Question: Assume that $X\_n$ is a sequence of a zero-mean and unit variance random variables (and maybe having density w.r.t. to Lebesgue). Can we conclude that $ P(X\_n \in [0,R\_n]) $ is bounded away from zero eventually, say $$\liminf\_{n \to \infty} P(X\_n \in [0,R\_n]) > 0$$ assuming that $R\_n \to \infty$. Intuitively, $X\_n$ should put some nonvanishing amount of mass on the positive real line because of the zero-mean assumption and the condition of unit variance should prevent that mass from escaping to infinity (?)

Here is a partial argument: We have
\begin{align\*}
P(X\_n \in [0,R\_n]) &= 1 - P(X\_n > R\_n) - P(X\_n < 0) \\
&\ge 1 - P(|X\_n| > R\_n) - P(X\_n < 0) \\
&\ge 1 - \frac{E X\_n^2}{R\_n^2} - P(X\_n < 0) \\
&= 1 - o(1) - P(X\_n < 0)
\end{align\*}
Thus, the problem reduces to arguing that $P(X\_n < 0)$ stays away from 1.

Answer: 
Here's a proof if a third moment condition is satisfied. 

Suppose that $\mathbb EX=0$, $\mathbb EX^2=1$ and $\mathbb E|X|^3\le K$. 
Then let $Z=|X|$. We use Cauchy-Schwarz: $Z^2=Z^{1/2}Z^{3/2}$, so that $(\mathbb EZ^2)^2\le \mathbb EZ\cdot \mathbb EZ^3$. This gives $\mathbb EZ\ge \frac 1K$. Hence $\mathbb EX\mathbf 1\_{X>0}\ge \frac{1}{2K}$. 

Also $\mathbb EX^2\mathbf 1\_{X>0}\le 1$. Now $\mathbb EX\mathbf 1\_{X>4K}\le (1/4K)\mathbb EX^2\mathbf 1\_{X>4K}\le 1/(4K)$, so that $\mathbb EX\mathbf 1\_{X\in [0,4K]}\ge 1/(4K)$ and $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$.
Filling in the details for Anthony's argument: 

Assume that $\mathbb E |X|^3 \le c$ for numerical constant $c > 0$.

Let $X^+ = X \mathbf 1\_{X > 0}$ and $X^- = (-X) \mathbf 1\_{X < 0}$. Then, $X = X^+ - X^-$. Let $\mu = \mathbb E X^+ = \mathbb E X^-$. Then, by Anthony's argument $2 \mu = \mathbb E |X| \ge 1/c$.

For $a \le \mu$, we have $\mathbb E X^+ \le a + \mathbb{E} X^+ \mathbb1\_{X^+ > a} \le a + [\mathbb E (X^+)^2]^{1/2} [\mathbb P(X^+ > a)]^{1/2} $ or
\begin{align}
 \mathbb P(X^+ > a) \ge \frac{(\mu-a)^2}{\mathbb E (X^+)^2} \ge (\mu-a)^2 \quad (\*)
\end{align}
using $\mathbb E (X^+)^2 \le \mathbb E X^2 \le 1$. Take $a = \frac1{4c}$ so that $a \le \frac1{2c} \le \mu$. Then, $\mathbb P(X^+ > \frac1{4c}) \ge \frac1{16c^2}$. Same bound holds for $X^-$ by symmetry. 

We can conclude that $X$ does not belong to $[-\frac{1}{4c},\frac{1}{4c}]$ with probability at least $1 -\frac{1}{8c^2}$.

EDIT: Apparently $(\*)$ is called [Paley-Zygmund inequality](https://en.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality) if one takes $a = \theta \mu$ for $\theta \in [0,1]$.