Let $M = n^2 + \frac{1}{2}n$.

Lemma 1. $B_n - A_n < \frac{1}{2}$.

Indeed,
$$(B_n - A_n) = \sum_{k=1}^{n} (\sqrt{n^2 + 2k} - \sqrt{n^2 + 2k - 1}) = \sum_{k=1}^{n} \frac{1}{\sqrt{n^2 + 2k} + \sqrt{n^2 + 2k - 1}} < \sum_{k=1}^{n} \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}$$
proving the lemma.

Lemma 2. $A_n < M < B_n$.

Proof. Observe that
$$(A_n - n^2) = \sum_{k=1}^{n} \left( \sqrt{n^2 + 2k - 1} - n \right) = \sum_{k=1}^{n} \frac{2k - 1}{\sqrt{n^2 + 2k - 1} + n} < \sum_{k=1}^{n} \frac{2k - 1}{n + n} = \frac{n^2}{2n} = \frac{n}{2}$$
as $\sum_{k=1}^{n}(2k-1) = n^2$, proving $A_n - n^2 < \frac{n}{2}$ or $A_n < M$. Similarly,
$$(B_n - n^2) = \sum_{k=1}^{n} (\sqrt{n^2 + 2k} - n) = \sum_{k=1}^{n} \frac{2k}{\sqrt{n^2 + 2k} + n} > \sum_{k=1}^{n} \frac{2k}{(n+1) + n} = \frac{n(n+1)}{2n+1} > \frac{n}{2}$$
as $\sum_{k=1}^{n}(2k) = n(n+1)$, so $B_n - n^2 > \frac{n}{2}$ hence $B_n > M$, as desired. $\square$

By Lemma 2, we see that $A_n$ and $B_n$ are positive real numbers containing $M$ between them. When $n$ is even, $M$ is an integer. This implies $\lfloor A_n \rfloor < M$, but $\lfloor B_n \rfloor \ge M$, which means we cannot have $\lfloor A_n \rfloor = \lfloor B_n \rfloor$.

When $n$ is odd, $M$ is a half-integer, and thus $M - \frac{1}{2}$ and $M + \frac{1}{2}$ are consecutive integers. So the above two lemmas imply
$$M - \frac{1}{2} < B_n - (B_n - A_n) = A_n < B_n = A_n + (B_n - A_n) < M + \frac{1}{2}.$$
This shows $\lfloor A_n \rfloor = \lfloor B_n \rfloor = M - \frac{1}{2}$.

Thus, the only integers $n \ge 3$ that satisfy the conditions are the odd numbers and all of them work. $\square$