The answer is no.
For a counterexample, let $T$ be the theory that, in addition to the assertions that the $a_j$'s are distinct, also asserts each sentence $\phi_{n,k,j}$, the sentence asserting that if all the $b_1,\ldots,b_n$ are among the $a_1,\ldots,a_k$, then $b_{\langle n,k\rangle}\neq a_j$, where $\langle n,k\rangle$ is a function on the natural numbers that is increasing in each coordinate and with $n,k\lt\langle n,k\rangle$.
This theory is consistent, since we can make all the $b_i$'s different from the $a_j$'s. Further, no model of $T$ has all the $b_i$'s among the $a_j$'s, since if $b_1$ is among the $a_j$'s, then some $b_{\langle 1,k\rangle}$ is not.
But meanwhile, there is no $n$ as you request. Fix any $n$ and let $k_1\gt\langle 1,1\rangle$ and $k_{i+1}$ be larger than $\langle i, k_i\rangle$. Now let $b_1,\ldots,b_n$ be among $a_j$ for $j$ above $k_n$, and all other $b_m$ not among the $a_r$. This will satisfy $T$, while having $b_1,\ldots,b_n$ among the $a_i$'s.
(Perhaps this idea can be simplified...)