Answer: $128$

With some initial trials, we can list the numbers of the students who get candies:
$1, 2, 4, 5, 10, 11, 13, 14, 28, 29, \ldots$

The base $3$ representation of the numbers are
$1, 2, 11, 12, 101, 102, 111, 112, 1001, 1002, \ldots$,
and if we subtract $1$ from each number, they become the following
$0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, \ldots$,
which are precisely those whose base $3$ representations consist of the digits $0$ and $1$ only. Indeed, we can easily prove by induction that student $n$ gets a candy if and only if the base $3$ representation of $n-1$ consists of the digits $0$ and $1$ only (call such a positive integer $n$ 'good'). This statement holds for the numbers we have listed. Now suppose $n$ is larger than $29$ and the statement holds for smaller $n$.

* Suppose $n$ is good. If student $n$ does not get a candy, there exist students $\ell$ and $m$ who both got candies (where $\ell < m < n$) and such that $\ell, m, n$ form an arithmetic sequence. Consider the numbers $\ell - 1, m - 1, n - 1$ (in base $3$, same for below). They all consist of the digits $0$ and $1$ only, and $2(m - 1) = (\ell - 1) + (n - 1)$. This is impossible, since each digit on the left is $0$ or $2$, while at least one digit on the right is $1$ (there is no carry in the addition, and there must be at least a place which gives $0+1$ since $\ell - 1 \neq n - 1$). This contradiction establishes the fact that student $n$ gets a candy.

* Suppose $n$ is not 'good'. Then $n - 1$ consists of at least one digit $2$. Suppose the place values of all the digits $2$ are $3^{k_1}, 3^{k_2}, \dots, 3^{k_d}$ respectively. Then student $n$ cannot get a candy since students $n - 2 \cdot (3^{k_1} + 3^{k_2} + \dots + 3^{k_d})$ and $n - (3^{k_1} + 3^{k_2} + \dots + 3^{k_d})$ both get candies. (For example, if $n = 74$ then $n - 1$ has base $3$ representation $2201$. Both students $2$ and $38$ got candies since the base $3$ representations of $1$ and $37$ are $1$ and $1101$ respectively. As $2, 38, 74$ form an arithmetic sequence, student $74$ cannot get a candy.)

This completes the induction. Note that $2023_{10} = 2202221_3$. So the largest $n$ which is 'good' has the property that the base $3$ representation of $n - 1$ is $1111111$. The answer is thus $2^7 = 128$.