If $b>2$, we define the sets $X=\left\{n\in\mathbb Z\mid n\in\left\{1,a-1\right\}\pmod b\right\}\cap \left\{1,2,\ldots,2024\right\}$ and $Y=\left\{1,2,\ldots,2024\right\}\setminus X$. Both sets $X$ and $Y$ are non-empty. Moreover, for every pair $\left(x,y\right)\in X\times Y$, we have $x+y\neq a$ and $|x-y|\neq b$. Consequently, it is impossible to construct a permutation of $\left\{1,2,\ldots,2024\right\}$ that satisfies the conditions. Hence, $b\in\left\{1,2\right\}$. If $b=1$, the identity permutation satisfies the conditions. If $b=2$, the permutation $1,3,\ldots,2023,2,4,\ldots,2024$ works.