Proof For $a \in (\sqrt{n}, \sqrt{2n}]$, $\text{lcm}(a, a+1) = a(a+1) > n$, so $|A \cap (\sqrt{n}, \sqrt{2n}]| \le \frac{1}{2}(\sqrt{2}-1)\sqrt{n} + 1$.
For $a \in (\sqrt{2n}, \sqrt{3n}]$, we have
$$\text{lcm}(a, a+1) = a(a+1) > n,$$
$$\text{lcm}(a+1, a+2) = (a+1)(a+2) > n,$$
$$\text{lcm}(a, a+2) \ge \frac{1}{2}a(a+2) > n.$$
So
$$|A \cap (\sqrt{2n}, \sqrt{3n}]| \le \frac{1}{3}(\sqrt{3}-\sqrt{2})\sqrt{n} + 1.$$
Similarly
$$|A \cap (\sqrt{3n}, 2\sqrt{n}]| \le \frac{1}{4}(\sqrt{4}-\sqrt{3})\sqrt{n} + 1.$$
Hence
$$\begin{aligned} |A \cap [1, 2\sqrt{n}]| & \le \sqrt{n} + \frac{1}{2}(\sqrt{2}-1)\sqrt{n} + \frac{1}{3}(\sqrt{3}-\sqrt{2})\sqrt{n} \\ & \quad + \frac{1}{4}(\sqrt{4}-\sqrt{3})\sqrt{n} + 3 \\ & = \left(1 + \frac{\sqrt{2}}{6} + \frac{\sqrt{3}}{12}\right)\sqrt{n} + 3. \end{aligned}$$
Let $k \in \mathbb{N}^*$, suppose $a, b \in (\frac{n}{k+1}, \frac{n}{k})$, $a > b$, and $\text{lcm}(a, b) = as = bt$, where $s, t \in \mathbb{N}^*$. Then
$$\frac{a}{(a, b)s} = \frac{b}{(a, b)t}.$$
Since $\text{gcd}(\frac{a}{(a, b)}, \frac{b}{(a, b)}) = 1$, so $\frac{b}{(a, b)}|s$. It follows that
$$\begin{align*} \operatorname{lcm}(a, b) &= as \ge \frac{ab}{(a, b)} \ge \frac{ab}{a-b} \\ &= b + \frac{b^2}{a-b} > \frac{n}{k+1} + \frac{\left(\frac{n}{k+1}\right)^2}{\frac{n}{k} - \frac{n}{k+1}} \\ &= n. \end{align*}$$
Therefore, $|A \cap (\frac{n}{k+1}, \frac{n}{k})| \le 1$.
Suppose $T \in \mathbb{N}^*$ such that $\frac{n}{T+1} \le 2\sqrt{n} < \frac{n}{T}$. Then
$$\begin{aligned} |A \cap (2\sqrt{n}, n]| &\le \sum_{k=1}^{T} \left|A \cap \left(\frac{n}{k+1}, \frac{n}{k}\right]\right| \\ &\le T < \frac{1}{2}\sqrt{n}. \end{aligned}$$
By the above arguments, we arrive at
$$|A| \le \left(\frac{3}{2} + \frac{1}{6}\sqrt{2} + \frac{1}{12}\sqrt{3}\right)\sqrt{n} + 3 < 1.9\sqrt{n} + 5.$$