Proposed by me, yay! Let $A = \left\lfloor \frac{1}{2x} \right\rfloor$ -- then the equation gives $x = \frac{2(n-A)}{n(n+1)}$ and now substituting in the definition of $A$ yields $$ A = \left\lfloor\frac{n(n+1)}{4(n-A)}\right\rfloor. $$The latter equality is a necessary and sufficient condition for the corresponding $x$ to be a solution to the equation. Let us also observe that $A$ is an integer and that $1 \leq A \leq n - 1$ for $n\geq 3$ -- indeed, if $A=0$, then $0 = \left\lfloor \frac{n+1}{4} \right\rfloor \geq 1$; if $A=n$, the right-hand side is undefined; and if $A<0$ or $A>n$, then the sides have different signs. a) For $n=8$ we want $A = \left\lfloor \frac{18}{8-A} \right\rfloor$. By the above, $A$ is an integer between $1$ and $7$ inclusive. A direct verification shows that only $A=3$ and $A=4$ are solutions, with the corresponding $x$ being $x=\frac{5}{36}$ and $x=\frac{1}{9}$. b) It suffices to have at least $2021$ integer solutions $1\leq A \leq n-1$ to $A \leq \frac{n(n+1)}{4(n-A)} < A + 1$ for some $n$. The left inequality is equivalent to $(2A - n)^2 + n \geq 0$ and holds for all $A$. The right inequality is equivalent to $(2A-n+1)^2 < n + 1$ and hence holds precisely for $\frac{n-1 - \sqrt{n+1}}{2} < A < \frac{n-1+\sqrt{n+1}}{2}$. Observe that this range for $A$ is tighter than $1 \leq A \leq n - 1$ for $n\geq 6$, as $(n-3)^2 > n + 1$ and $(n-1)^2 > (n+1)$ for these $n$. Finally, the difference between the endpoints of the interval $\left(\frac{n-1 - \sqrt{n+1}}{2}, \frac{n-1+\sqrt{n+1}}{2}\right)$ is $\sqrt{n+1}$ and hence for sufficiently large $n$ this interval must contain at least $2021$ integers. This completes the proof.

Is there a solution using the expression $[1/2x]=1/2x-${$1/2x$}?

a) is trivial, so I'm not going to bother typing my solution to it, I'm just going to show my in contest sol to b), because it's different to @Assassino9931 's solution: We're looking for just one $n$, so let $n=2a$ be even (so we can divide both sides by $2$, no other reason). Then let's transform the equation into: $$2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor-n+1=(n+1)(1-nx)\Longrightarrow 2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor-2a+1=(2a+1)(1-2ax)$$$$\Longrightarrow 2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor+2a(2a+1)x=4a\Longrightarrow \Bigg\lfloor \frac{1}{2x}\Bigg\rfloor+a(2a+1)x=2a$$Now the only logical thing to do is set $a(2a+1)x=a+c$ and $\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor=a-c$, where $c$ is a constant. Then we express $x=\frac{a+c}{a(2a+1)}\in \mathbb{R}$ and we want to prove that: $$a-c\overset{(1)}{\leq} \frac{a(2a+1)}{2(a+c)}=\frac{1}{2x}\overset{(2)}{<}a-c+1$$Inequality $(1)$ is always true as $a+c>0$ and obviously $-2c^2<a$ and the second inequality is equivalent to $2c^2-2c<a$, but $c$ is a constant and so for every $a>2\times 2021^2-4042$ we have that $\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor=a-c$, for every $c=1,2,\ldots ,2021$, so for every number $n>2(2\times 2021^2 - 4042)$ the equation has at least $2021$ different solutions as all $x=\frac{a+c}{a(2a+1)}$ are obviously different for $c=1,2,\ldots , 2021$. Therefore we've proven b)

Marinchoo wrote: a) is trivial, so I'm not going to bother typing my solution to it, I'm just going to show my in contest sol to b), because it's different to @Assassino9931 's solution: We're looking for just one $n$, so let $n=2a$ be even (so we can divide both sides by $2$, no other reason). Then let's transform the equation into: $$2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor-n+1=(n+1)(1-nx)\Longrightarrow 2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor-2a+1=(2a+1)(1-2ax)$$$$\Longrightarrow 2\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor+2a(2a+1)x=4a\Longrightarrow \Bigg\lfloor \frac{1}{2x}\Bigg\rfloor+a(2a+1)x=2a$$Now the only logical thing to do is set $a(2a+1)x=a+c$ and $\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor=a-c$, where $c$ is a constant. Then we express $x=\frac{a+c}{a(2a+1)}\in \mathbb{R}$ and we want to prove that: $$a-c\overset{(1)}{\leq} \frac{a(2a+1)}{2(a+c)}=\frac{1}{2x}\overset{(2)}{<}a-c+1$$Inequality $(1)$ is always true as $a+c>0$ and obviously $-2c^2<a$ and the second inequality is equivalent to $2c^2-2c<a$, but $c$ is a constant and so for every $a>2\times 2021^2-4042$ we have that $\Bigg\lfloor \frac{1}{2x}\Bigg\rfloor=a-c$, for every $c=1,2,\ldots ,2021$, so for every number $n>2(2\times 2021^2 - 4042)$ the equation has at least $2021$ different solutions as all $x=\frac{a+c}{a(2a+1)}$ are obviously different for $c=1,2,\ldots , 2021$. Therefore we've proven b) Exactly same with mine that I did in exam and I think it's much more better than offical solution. Official solution is as @Assassino9931.

Lukaluce wrote: Let $n$ ($n \ge 1$) be an integer. Consider the equation $2\cdot \lfloor{\frac{1}{2x}}\rfloor - n + 1 = (n + 1)(1 - nx)$, where $x$ is the unknown real variable. (a) Solve the equation for $n = 8$. (b) Prove that there exists an integer $n$ for which the equation has at least $2021$ solutions. (For any real number $y$ by $\lfloor{y} \rfloor$ we denote the largest integer $m$ such that $m \le y$.) $2n-2\cdot \lfloor{\frac{1}{2x}}\rfloor=n(n+1)x$. If $x<=0$ then $LHS>0$ and $RHS<0$ contradiction. If $x>=1$ then $x=2/(n+1)$ contradiction.(we can suppose that $n>=2$) So $0<x<1$ let $y$ natural number such that:$1/y>x>1/(y+1)$ We have two cases if $y=2k$ and if $y=2k+1$ in both of them we have$2\cdot \lfloor{\frac{1}{2x}}\rfloor=2k$ We condition we have $x=(16-2k)/72$ and $1/y>x>1/(y+1)$ after solving the inequality we have $x=5/36$. If $x=1/y$ we can get $y=9$. b) $x=(2n-2k)/n(n+1)$ with $1/y>x>1/(y+1)$ and $y=2k+1$ so we have: $1/(2k+1)>(2n-2k)/n(n+1)>1/2k+2$ After solving both of them we have: The LHS gives $k>n/2$ or $k<(n-1)/2$ And the right (2n-2-√(10n))/4<k<(2n+2+√(10n))/4 For enough big $n$ we can find at least $2021$ natural number equal to $k$. Let me know if I am wrong.

For part a) Rearranging gives $8-\Bigg \lfloor \frac{1}{2x} \Bigg \rfloor = 36x$ which means $36x=a$ is an integer. We have $8=a+ \Bigg \lfloor \frac{18}{a} \Bigg \rfloor$. It is easy to see $a>0$ and rest is casework. (We have $a \le 8$ obviously and...)

me and mathmax12 could both solve part(a) with the same solutions: if $n=8$, then $2\cdot \lfloor{\frac{1}{2x}}\rfloor - 7 = 9 - 81x$, so $2\cdot \lfloor{\frac{1}{2x}}\rfloor = 16 - 81x$, now we can bash out cases to get that $x=\boxed{\frac{5}{36} and \frac{1}{9}}.$