Sum of boolean random variables with small correlations. Suppose $X_1,\dots,X_N$ are $0/1$-random variables  and the following information is known:
$$
\Pr[X_i=1]\geq 1/n, ~ \Pr[X_iX_j=1]\leq 100/n^2.
$$
Let $X=\sum_{i=1}^N X_i$. Assume $N=n^{100}$ for example. 
As $N\gg n$, if $X_i$'s were pairwise independent then we have
$\Pr[X>0]\approx 1$ using e.g. Chebyshev's inequality.
Here $X_i$'s are not pairwise independent, but the correlation of each two of them is $O(1/n)$. I am wondering if $\Pr[X>0]\approx 1$ still holds, and if so, how can one show it? I've tried Chebyshev's inequality:
$$
\Pr\left[X=0\right]\leq \frac{\mathbf{Var}[X]}{\mathbf{E}[X]^2}
$$
but it does not yield any non-trivial bound as $\mathbf{Var}[X]\geq \mathbf{E}[X]^2$. One can use Cauchy-Schwarz inequality to show
$$
\Pr[X>0]=\mathbf{E}\left[\mathbf{1}^2_{X>0}\right]\geq \frac{\left(\mathbf{E}[\mathbf{1}^2_{X>0}\cdot X]\right)^2}{\mathbf{E}\left[X^2\right]}=\frac{\mathbf{E}[X]^2}{\mathbf{E}\left[X^2\right]}\geq 1/100-o(1)
$$
which is still not satisfying (here $\mathbf{1}_{X>0}$ is the indicator random variable for the event $X>0$). Is there any inequalities in probability theory that can be applied here to show $\Pr[X>0]\approx 1$? Thanks! 

Here is a counterexample. 
Assume that $(Y_i)_i$ is i.i.d. Bernoulli 0-1 with $\mathbb E(Y_i)=y/n$, and that $Z$ is Bernoulli 0-1 and independent of $(Y_i)_i$ with $\mathbb E(Z)=z$. Let $$X_i=ZY_i.$$ Then $\mathbb E(X_i)=\mathbb E(Z)\mathbb E(Y_i)=zy/n$ and $\mathbb E(X_iX_j)=\mathbb E(Z)\mathbb E(Y_i)\mathbb E(Y_j)=zy^2/n^2$ hence the hypothesis on the moments hold if $zy\geqslant1$ and $zy^2\leqslant100$ (say, $z=1/2$ and $y=3$ for $n\geqslant3$) but $\mathbb P(X=0)\geqslant\mathbb P(Z=0)=1-z\gt0$ if $z\lt1$, irrespectively of the value of $N$.
