Mongolian Mathematical Olympiad

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 $n$) $\mapsto ka$ (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 $$\{a_1,a_2,\dots ,a_{n-2}(\mathrm{mod}n)\}.$$ 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 $(k^2-1)b\equiv 0(\mathrm{mod}n)$ for some $b\in B$ satisfying $b\ne 0(\mathrm{mod}n)$ 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 $kb\equiv b(\mathrm{mod}n)$ and $B=\{b,kb\}$ if $kb\ne b(\mathrm{mod}n)$. 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.