Answer: $92$.
Set $n = 2023 = 7 \times 17^2$ and denote by $\mathbb{Z}/n\mathbb{Z}$ the set of all residues modulo $n$. For any natural number $k$, denote by $k: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ the map defined by $a$ (mod $n$) $\mapsto k a$ (mod $n$).
Assume that $k$ is nice and $a_1, a_2, \dots, a_{n-2}$ are integers satisfying the two conditions. Then $A \subseteq kA$ by the condition (2), where $A$ denotes the set
$$\{a_1, a_2, \dots, a_{n-2} \pmod{n}\}.$$
Therefore, $|A| = |kA| = n-2$ by the condition 1 and so $A = kA$. Since $kA \subseteq k\mathbb{Z}/n\mathbb{Z}$ and $|k\mathbb{Z}/n\mathbb{Z}| = n/(n,k)$ we have $(n,k) = 1$, whence the map $k: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ is invertible. Thus, by setting $B = \mathbb{Z}/n\mathbb{Z} \setminus A$, we get $kB = B$ and $|B| = 2$. More precisely, if $B = \{b_1, b_2\}$ then $(k b_1, k b_2) = (b_1, b_2)$ or $(k b_1, k b_2) = (b_2, b_1)$. Thus $(k^2-1) b \equiv 0 \pmod{n}$ for some $b \in B$ satisfying $b \ne 0 \pmod{n}$ and so $(k^2-1, n) \ne 1$.
Now assume that $(k, n) = 1 \ne (k^2 - 1, n)$. By setting $b = n/(k^2 - 1, n)$, we choose $B = \{0, b\}$ if $k b \equiv b \pmod{n}$ and $B = \{b, k b\}$ if $k b \ne b \pmod{n}$. Then the elements of $A = \mathbb{Z}/n\mathbb{Z} \setminus B$ satisfy the two given conditions.
Thus, it suffices to find the largest even integer $k$ such that $(k, n) = 1 \ne (k^2 - 1, n)$ and $k \le 100$. It is now straightforward to check that $92$ is nice while $100$, $98$, $96$, $94$ are not.