well basically the problem is to count the numbers from $$$1$$$ to $$$n$$$ that have odd number of divisors.

if you think about divisors of a number, you can see that for every divisor $$$d$$$ of a number $$$x$$$, there's its let's say mirror divisor $$$x/d$$$. So you may guess that every number actually should have even number of divisors (if for every $$$d$$$ you have $$$x/d$$$ you should end up with even number of divisors right?). But it's not the case. In which cases a number has odd number of divisors then? When $$$d = x/d$$$, since you only count number of unique divisors (you don't count the same number twice).

Now the problem is to find number of such $$$i$$$ that there's some divisor $$$d = i/d$$$. Simple math and you see that $$$d^2 = i$$$, so $$$i$$$ should be perfect square. You can count them manually, since the square root of the biggest perfect square $$$i$$$ cannot be bigger than $$$d^2 = i \leq n, d \leq \sqrt{n}$$$ (complexity $$$O(\sqrt{n})$$$)

And finally you can notice that integer $$$\sqrt{n}$$$ is literally the biggest such number $$$d$$$, and every $$$d$$$ less than $$$\sqrt{n}$$$ works as well (if $$$d^2 \leq n$$$, then obviously $$$(d-1)^2 \leq n$$$ if $$$d > 0$$$). So the answer is $$$\sqrt{n}$$$

I don't see the reason why it is so disliked

I solved this problem like the second I read the statement, it's all about practice I guess then afterall. Solve more problems like this and you'll get the idea how to think to solve this kind of problems.