Prove that there exists a natural number $n$ that has more than $2017$ divisors $d$ satisfying $\sqrt{n} \le d < 1{,}01\sqrt{n}$ 
Prove that there exists a natural number $n$ that has more than $2017$ divisors $d$ satisfying
$$\tag 1 \sqrt n \le d < 1.01\sqrt n$$

My reasoning was that $\tag 2 1.01\sqrt n-\sqrt n\ge2019$ must occur. Otherwise, there won't be enough space to fit in enough numbers in $(1)$. $(2)$ gives us that $n\geq 201900^2$. We also have that $201900^2<201900!$ so we can pick any factorial greater than or equal to $201900!$ and that satisfies the desired condition. Is this solution correct?
 A: Let
$$
n=(10^6-1)^{4032}(10^6+1)^{4032}.
$$
Then the numbers
$$
s_j=(10^6-1)^j(10^6+1)^{4032-j},
$$
$j=0,1,2,\ldots,2016$, are clearly all factors of $n$.
We also have
$$s_0>s_1>\cdots >s_{2016}=\sqrt n,$$
so they are distinct and satisfy the lower bound.
On the other hand
$$
\frac{s_0}{s_{2016}}= \frac{(10^6+1)^{2016}}{(10^6-1)^{2016}}<1.01\qquad(*)
$$
so we have $s_j<1.01\sqrt n$ for all $j$ also.
I checked $(*)$ with Mathematica, but elementary estimates will also get the job done. At least if we replace $10^6$ by something bigger.
A: Generally speaking, I expect that such a contest problem can be done via induction, and we just need to push through.
Define a divisor that satisfies those conditions to be "valid".
Base case: $D = 1$, $n_1=1$ suffices.
Induction step: Suppose it is true for some $D$ with corresponding $n_D$.
Consider $n_{D+1} = p^2 n_{D}$ where $p$ is to be determined below.
Then, for any valid divisor $d_D$ of $n_D$, we have that the corresponding $ pd_D$ satisfies $pd_D \mid n_{D+1}$ and $\sqrt{n_{D+1}} \leq pd_{D} < 1.01 \sqrt{n_{D+1}}$. Hence this gives us (at least) $ D$ valid divisors.
How can we force another distinct valid divisor?
If we have an integer $p \in [ \sqrt{n_D}, 1.01 \sqrt{n_D}) $ that is not a divisor of $n_D$, then we could use $ n_{D+1} = p n_D    $.
If not, look at the interval $ [ \sqrt{ 4^k  n_D } ,   1.01 \sqrt{4^k n_D})$, and show that we will eventually find such a $p$ that is not a divisor of $4^k n_D$. (Try proving without any high power theorem). Then, we can use $n_{D+1} = (2^k p)^2 n_D$.
