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! 

This is a comment too long to fit in the usual format.
Note that
$$
1 \leq Pr(X_1=0)+Pr(X_2=0)+Pr(X_1=X_2=1) \leq \frac{1}{n}+\frac{1}{n}+\frac{100}{n^2}=\frac{2n+100}{n^2},
$$
we deduce that $n^2-2n-100 \leq 0$, so $n \leq 11$. 
So you cannot make $n$ as large as you want.
