If $A$ is a perfect square, i.e. there exists $B \in \mathbb{N}$ such that $A = B^2$, then $(A+k)^2 - A = (B^2+k)^2 - B^2 = (B^2+B+k)(B^2-B+k)$ for all $k = 1, n$, and exactly one of the (consecutive) numbers $B^2+B+1, B^2+B+2, \dots, B^2+B+n$ is a multiple of $n$.

Conversely, if $A$ is not a perfect square, then it has a prime factor that occurs in the prime factorization of $A$ at an odd exponent. Let $p$ be such a prime and $j \in \mathbb{N}$ such that $p^{2j-1} \mid A$, but $p^{2j} \nmid A$. We choose $n = p^{2j} \in \mathbb{N}$ and show that none of the numbers $(A+1)^2-A, (A+2)^2-A, (A+3)^2-A, \dots, (A+n)^2-A$ is a multiple of $n$. Indeed, if $n \mid (A+m)^2-A$, for some $m \in \{1, 2, \dots, n\}$, i.e. $p^{2j} \mid A^2 + 2Am + m^2 - A$, from $p^{2j-1} \mid A$ it follows that $p^{2j-1} \mid m^2$, hence $p^j \mid m$. But then $p^{2j} \mid (A+m)^2$ and $p^{2j} \mid (A+m)^2 - A$, hence $p^{2j} \mid A$, which is a contradiction.