How to Solve Problem Similar to IMO(1995) Problem Two years ago conjecture,Today I again remind of the problem

(conjecture) Let $ n$ be an postive integer number. How many $ n$-element subsets $A$ of $ \{1,2,\dots,2n\}$ are there such that $1+2+\cdots+2n$ is divisible by the sum of the elements of $A$.

I have found this similar problem:IMO (1995) last problem:IMO
two year ago idea:
let $$A=\{x_{1},x_{2},\cdots,x_{n}\},1\le x_{1}&lt;x_{2}&lt;\cdots&lt;x_{n}\le 2n$$
and we have
$$1+2+3+\cdots+2n\equiv 0\pmod{x_{1}+x_{2}+\cdots+x_{n}}$$
Today ADD it,  By Now I have read Problem and from the book Chapter 7 example 4 (simaler problem):
(Solved )  Let $f(n)$ be the number of subsets of  $1,2,3,\cdots,n$ whose elements sum  to 0 $\pmod n$,the empty set is included,having the element sum equal to zero,Prove that
$$f(n)=\dfrac{1}{n}\sum_{d|n,d~odd}\varphi(d)2^{\frac{n}{d}}$$
Solution: Let
$$g(X)=\prod_{i=1}^{n}(1+X^i)=\sum_{k\ge 0}a_{k}X^k$$
if let $\varepsilon =e^{\frac{2i\pi}{n} }$,it is clear
$$f(n)=\sum_{j\ge 0}a_{jn}$$
and other hand easily be computed
$$\sum_{j=1}^{n}g(\varepsilon ^j)=n\sum_{j\ge 0}a_{jn}$$
if let $d=\dfrac{n}{\gcd{(n,j)}}$,and use well known
$$x^d-1=(x-\varepsilon^j)(x-\varepsilon^{2j})\cdot (x-\varepsilon^{dj})$$
so we have
$$(1+\varepsilon^j)(1+\varepsilon^{2j})\cdot (1+\varepsilon^{dj})=2$$
this shows that
$$g(\varepsilon ^j)=2^{n/d},d ~~is odd$$
and $0$ otherwise.
so
$$\dfrac{1}{n}\sum_{j=1}^{n}g(\varepsilon ^j)=\dfrac{1}{n}\sum_{d|n,d~odd}\varphi(d)2^{\frac{n}{d}}$$
if $n=p$ be prime,then it's 1995 IMO problem.
But for conjecture Now I don't solve it by now!
Hint:  Note that $\sum_{i=1}^{2n}i = n(2n+1)$  The sum of an $n$ element subset of $A$ is at least $\frac 12n(n+1)$, so the only sums that can divide $n(2n+1)$ are  $\frac 13n(2n+1)$, $\frac 12n(2n+1)$ and $n(2n+1)$.  You need to find how many $n$ element subsets sum to one of these.  The first and third may not be a whole number, and should be ignored if so.
$\frac {n(2n+1)}{\frac {n(n+1)}{2}}&lt;4$ ,so  equation which a sum divided is 2 or 3.
(i)When sum of two elements is 2n+1,  $$2n+1=(1,2n)(2,2n-1)\ldots(n,n+1)$$n pairs.By choosing $\frac n2$, $_nC_{\frac n2}$ pairs.This sum is just half of equation.

(ii)When sum of two elements is n or 3n$$(1,n-1)(2, n-2)\dots(\frac n2-1,\frac n2+1)$$
$$(n,2n)(n+1,2n-1)\ldots(\frac {3n}2-1,\frac {3n}2+1)$$This sum are $_\frac {n-2}2C_k$,$_\frac {n-2}2C_\frac {n-2k}2$ by choosing k,$\frac {n-2k}2$ .
These sum is$$\tag{1}(n*k)(3n*(\frac n2-k))$$
By the way,there is no common diviser of $(n,2n+1)$ and (1) is not diviser of equation whatever k is.

(iii)When sum is 3,
$(2n-2,1,2)(2n-6,3,4)\ldots(\frac {2n}3+1,\frac {2n}3-1,\frac {2n}3)$

$\frac {n}3+1$ pairs,$_{\frac n3+1}C_\frac n3=\frac n3+1 $

So answers are,using $_{\frac n3}C_{\frac n3}=1$

if $n=6k-2,6k-4,$ $_nC_{\frac n2}+1$

if $n=6k-1,6k-5$, $_nC_{\frac {n-1}2}+1$

if $n=6k$, $_nC_{\frac n2}+{\frac n3}+1 $

if $n=6k-3$,$_nC_{\frac {n-1}2}+{\frac n3}+1$