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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 fixed $k$ 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.

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.

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 $\phi_{k,j}$, asserting that if $b_1$ is among $a_1,\ldots,a_k$, then $b_{k+1}\neq a_j$.

This theory is consistent, since we can let every $b_k$ different from every $a_j$. Also, in any model of $T$, if $b_1=a_k$, then $b_{k+1}\neq a_j$ for any $j$. So it has the property that no model of $T$ has only 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, so the desired property fails.

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 fixed $k$ 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.

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...)

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 $\phi_{k,j}$, asserting that if $b_1$ is among  $a_1,\ldots,a_k$, then $b_{k+1}\neq a_j$.

This theory is consistent, since we can let every $b_k$ different from every $a_j$. Also, in any model of $T$, if $b_1=a_k$, then $b_{k+1}\neq a_j$ for any $j$. So it has the property that no model of $T$ has only 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, so the desired property fails.