Testing the
$$\frac{\lfloor \sqrt{n}\rfloor(1 + \lfloor\sqrt{n}\rfloor)}{2 n} = \frac{\lfloor \sqrt{k}\rfloor(1 + \lfloor\sqrt{k}\rfloor)}{2 k}$$
And found that for Oblong numbers (numbers of form $(n+1)n$) this equation is true. In other words this equation has infint solutions for those numbers. While for other ones just has discovered that it has finite solutions (experiment).
Trying to prove the last statement.
Let $t$ be a fixed number not equal to $1$.For $$\frac{\lfloor \sqrt{n}\rfloor(1 + \lfloor\sqrt{n}\rfloor)}{ n} = t$$ we require there to be an integer $k$ such that $$k^2\le n<(k+1)^2, k(k+1)=nt.$$ Then $$\frac{k}{k+1}\le\frac{1}{t}<\frac{k+1}{k}.$$
As $k$ tends to infinity, $\frac{k}{k+1}$ and $\frac{k+1}{k}$ both tend to $1$ and so only finitely many $k$ can satisfy the equations. Therefore there are only finitely many $n=\frac{k(k+1)}{t}$.