This is a pretty interesting question. Douglas S. Stones' answer provides an example that works (but has diameter 3). On the other hand, large diameter forces many induced subgraphs for all small values of $i$.
Lemma. Let $G$ be a graph with diameter $d \geq 8$. Then for all $4 \leq i \leq d/2$, the number of induced subgraphs of $G$ with $i$ vertices is more than $i$.
Proof. Let $P$ be an induced path of $G$ with $d$ vertices. Let $F_i$ be the set of all forests on $i$ vertices with maximum degree 2. Observe that $P$ contains all graphs in $F_i$ for all $i \leq d/2$ as induced subgraphs. Since $F_i > i$ for all $i \geq 4$, we are done.
The property seems harder to satisfy for larger values of $i$ which leads us to the following (updated) 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 $G$ has diameter at most 7.