This is a pretty interesting question.  Here are some trivial observations.  For an example of a connected non-bipartite graph that satisfies the property for $i=4$, we may take any odd cycle of length at least 5.  On the other hand, large induced cycles force many induced subgraphs for all small values of $i$.  

**Lemma.** Let $G$ be a graph with an induced cycle of length $g \geq 8$.  If $G$ is not a cycle, then for all $4 \leq i \leq g/2$, the number of induced subgraphs of $G$ with $i$ vertices is more than $i$.

**Proof.** Let $C$ be an induced cycle of $G$ of length $g$.  Let $F_i$ be the set of all forests on $i$ vertices with maximum degree 2.  Observe that $C$ contains all graphs in $F_i$ for all $i \leq g/2$.  Note that $|F_i|>i$ for all $i>4$.  This almost proves the lemma, except that $|F_4|=4$.  But here we use the hypothesis that $G$ is not a cycle, in which case $G$ contains an induced subgraph $H$ on 4 vertices with maximum degree at least 3.  Hence $H \notin F_4$ and we are done.  

The property seems harder to satisfy for larger values of $i$ which leads us to the following rash conjecture.

**Rash Conjecture.** Let $G$ be a connected, non-bipartite graph on $n$ vertices whose complement is also connected and non-bipartite.  If $G$ has at most $i$ induced $i$-subgraphs for some $3 < i < n/2$, then $G$ or its complement is an odd-cycle.