The answer is no.
Here is a simplified counterexample (my earlier post was more complicated).
Let $T$ be the theory asserting $a_i\neq a_j$ and also the assertions, for every particular $k$ and $j$, that if $b_1=a_k$, then $b_{k+1}\neq a_j$.
This theory is consistent, since we can let $b_1\neq a_k$ for any $k$, which vacuously satisfies all the implications. Also, in any model of $T$, if $b_1=a_k$, then $b_{k+1}\neq a_j$ for any $j$. So no model of $T$ consists only of the $a_i$'s.
But meanwhile, there is no $n$ as you request, since for any $n$, we can let $b_1=\cdots=b_n=a_n$, and $b_{n+1}\neq a_j$ any $j$. This will be a model of $T$ where the $b_1,\ldots,b_n$ are among the $a_i$'s, and so the requested property is violated.