Let $n$ be even, $n = 2k$. Then $\frac{n^2}{4} = k^2$ is even, so $\lfloor \frac{n^2}{4} \rfloor = k^2$ is a perfect square.

If $n$ is odd it can be written as $n = 2k + 1$, $k \ge 0$. In this case
$$\lfloor \frac{n^2}{4} \rfloor = \lfloor \frac{4k^2 + 4k + 1}{4} \rfloor = \lfloor k^2 + k + \frac{1}{4} \rfloor = k^2 + k = k(k + 1).$$
For $k(k + 1)$ to be a perfect square both $k$ and $k + 1$ have to be perfect squares since they are coprime. Hence, $k = a^2$ and $k + 1 = b^2$. This implies $(b-a)(b+a) = 1$. There are two possibilities, $b-a = 1 = b+a$ or $b-a = -1 = b+a$. Both imply $k=0$, so $n=1$ is the only odd number with the required property.

There is another way to deal with the case where $k^2 + k$ is a perfect square. The quadratic equation $k^2 + k - x^2 = 0$ has two solutions, $k_{1,2} = \frac{-1 \pm \sqrt{1+4x^2}}{2}$, so $1 + 4x^2 = y^2$ or, equivalently, $(y - 2x)(y + 2x) = 1$. This implies $x = 0$, so $k = 0$.

We conclude that $\lfloor \frac{n^2}{4} \rfloor$ is a perfect square if and only if $n = 1$ or $n$ is even.