a) Let $A = \{1, 2, \dots, n\}$ and $n = k^2$ for some positive integer $k$. Denote $B$ as the set of first $m$ odd positive integers. It's clear that $B$ is nice. Assume that $n \ge 2m-1$, let $C = A \setminus B$. $C$ is nice if and only if
$$\frac{(1 + 2 + \cdots + n) - m^2}{n - m} = n - m \iff n = n^2 - 4mn + 4m^2 \iff m = \frac{n-k}{2}.$$
Therefore, we can choose $m$ so that $C$ is nice. Hence, $n$ is amazing.

b) We prove that if $n = 4k + 2$ for some positive integer $k$, then $n$ is not amazing. Assume that $A = \{1, 2, \dots, n\}$ can be partitioned into nice subsets $A_1, A_2, \dots, A_m$ with cardinality $a_1, a_2, \dots, a_m$ respectively. Therefore,
$$\frac{n(n+1)}{2} = \sum_{x \in A} x = \sum_{i=1}^{m} \sum_{x \in A_i} x = \sum_{i=1}^{m} a_i^2 = \sum_{i=1}^{m} a_i = n \pmod{2},$$
that yields contradiction. $\square$