Does there always exist an even $m$ that is a multiple of exactly $n$ of the numbers $1$, $2$, ..., $2n$? Let $n>1$ be a positive integer. Then there exists a positive integer $m$ such that exactly half of the numbers $1$, $2$, $\ldots$, $2n$ divides $m$: one can take $m = (2n-1)!! = (2n-1) \times (2n-3) \times \cdots \times 3 \times 1$. 
My question is whether there also always exists an even number $m$ with this property. 
For instance, for $n=10$ we can take $m=144$ since of the numbers $1$, $2$, $\ldots$, $20$, exactly $10$ are divisors of $144$:
$$
 {\color{red}1} \;\; {\color{red}2} \;\; {\color{red}3} \;\; {\color{red}4} \;\; 5 \;\; {\color{red}6} \;\; 7 \;\; {\color{red}8} \;\; {\color{red}9} \;\; 10 \;\; 11 \;\; {\color{red}1\color{red}2} \;\; 13 \;\; 14 \;\; 15 \;\; \color{red}1\color{red}6 \;\; 17 \;\; \color{red}1\color{red}8 \;\; 19 \;\; 20 
$$
Thus, the question is:

Given a positive integer $n>1$, does there always exist an even positive integer $m$ such that of the numbers $1$, $2$, $\ldots$, $2n$ exactly $n$ are divisors of $m$?

A computer search shows that this is true for $n \leq 1000$ and I suspect that it is true in general. I would be interested to know whether someone has any suggestions or references for this problem.
 A: This is a proof for $n\ge93$.
Let $A_p=\{1\le k\le 2n:p|k\}$. For primes $p,q>\sqrt{2n}$, $A_p\cap A_q=\emptyset$ because $pq>2n$.
If we find some nice set of primes $Q=\{q_1,q_2,\cdots, q_k\}$ ($\sqrt{2n}<q_1< q_2<\cdots<q_k\le2n$) that $\displaystyle\sum_{t=1}^k|A_{q_t}|=n$, setting $\displaystyle m=\prod_{\substack{p\le 2n\\p\notin Q}}p^{\left[\log_p 2n\right]}$ works.
To show that such set $Q$ exists, we use two lemmas.

Lemma 1. For $x>185$, $$\sum_{p>\sqrt{x}}\left[\frac xp\right]>\frac x2$$
Proof)
From this, we have $$\left|\sum_{p
\le x}\frac1p-\log\log x-B\right|\le\frac1{10\log^2x}+\frac4{15\log^3x}$$ for $x\ge10372$. And according to wikipedia, $\pi(x)<1.3\frac{x}{\log x}$ for $x\ge17$. Using this we can get, $$\sum_{p>\sqrt{x}}\left[\frac xp\right]>\sum_{p>\sqrt{x}}\frac xp-\pi(x)\ge x\log2-1.3\frac{x}{\log x}-\frac{15}{2\log^2x}-\frac{12}{5\log^3x}>\frac x2$$And computation gives this lemma true for $x>185$.
Lemma 2. For $n\ge49$, $$\sqrt{2n}\le\pi(2n)-\pi(n)$$
Proof)
This is just a modification of Erdos's proof of Bertrand's postulate.
$$\frac{4^n}{2n}\le\binom{2n}{n}<(2n)^\sqrt{2n}4^{2n/3}(2n)^{\pi(2n)-\pi(n)}$$
Applying logarithm on both sides,
$$\pi(2n)-\pi(n)\ge \frac{\log 4}{3\log(2n)}n -\sqrt{2n}-1$$
For $n>5000$, $\frac{\log 4}{13}\sqrt {2n}>\log 2n$ holds. So we can get
$$\pi(2n)-\pi(n)\ge \frac{\log 4}{3\log(2n)}n -\sqrt{2n}-1>\frac 76\sqrt{2n}-1>\sqrt{2n}$$ for all $n>5000$.
Computation shows that lemma 2 is also true for $49\le n\le5000$.

Let $Q_0$ be the set of all primes $p$ in the range $\sqrt {2n}<p\le 2n$. Then we have $\displaystyle\sum_{p\in Q_0}|A_p|>n$ from lemma 1. Now we do a process of discarding $p$ from $Q_0$ starting from the smallest element of $Q_0$. Let $q_s$ as the smallest element of $Q_s$ and $Q_{l+1}:=Q_l-\{q_l\}$. Then there exist an $s$ that $$\sum_{p\in Q_{s+1}}|A_p|< n\le\sum_{p\in Q_s}|A_p|$$
Since $$\sum_{p\in Q_s}|A_p|-n< |A_{q_s}|=\left[\frac{2n}{q_s}\right]\le\sqrt{2n}\le\pi(2n)-\pi(n)$$ from lemma 2, we can pick $\displaystyle\sum_{p\in Q_s}|A_p|-n$ primes in the range $n<p\le 2n$ given that $q_s\le n$. If we put $Q$ as those primes discarded from $Q_s$, we have $\displaystyle\sum_{p\in Q}|A_p|=n$ we are done. So it only remains to show $q_s\le n$.
Assume that $q_s>n$. Then we have $$n\le\sum_{p\in Q_s}|A_p|=\sum_{p\in Q_s}1=|Q_s|\le\pi(2n)-\pi(n)=\pi(2n-1)-\pi(n)<n$$ which is a contradiction.
Remark. Note that arguments made in this proof except for lemma 1 are elementary. Lemma 1 can also be shown for sufficiently large $x$ by the Merten's theorem but it is not effective bound because my calculation gave it only being true for roughly $x>10^{30}$...
To wrap it up, we can prove the statement by elementary means for all sufficiently large $n$ but to give the complete proof using this argument, we need a powerful effective bound.
