The answer is negative:  It is possible that there is no good choice of $i,j$.
Let $T$ be a uniform spanning tree in the infinite ladder ${\bf Z} \times \{0,1\}$. To be precise, this is a weak limit of uniform spanning trees from finite
ladders as shown to exist in [1]. See Chapter 4 of [2] for more information.
Let $A_i$ be the event that the $i$'th rung of the ladder is an edge of $T$.
Then ${\bf P}(A_i)$ does not depend on $i$ and ${\bf P}(A_i \cap A_j) < {\bf P}(A_i) {\bf P}(A_i)$ because the transfer-current matrix has positive entries.
Now ${\bf P}(A_i)>1/2$, but this is easily addressed by replacing $A_i$ with the intersection $A_i \cap \{B_i=1\}$,   where $\{B_i\}$ are i.i.d.\ fair coins independent of the events $\{A_i\}$.

Many other examples of stationary determinantal processes that can be used as counterexamples to your question are discussed in [3].  

[1] Pemantle, R. (1991), Choosing a spanning tree for the integer lattice uniformly. Ann. Probab., 19(4), 1559–1574.

[2] Lyons, R.  and   Peres, Y. (2016), Probability on Trees and Networks,
 Cambridge University Press, to appear. Available at http://pages.iu.edu/~rdlyons

[3]   Lyons, R. and Steif, J.E. (2003) Stationary determinantal processes: phase multiplicity, Bernoullicity, entropy, and domination. Duke
Math. J., 120(3), 515–575. http://pages.iu.edu/~rdlyons/pdf/dyn.pdf