The maximum $n$, its smallest $x$ and the equality now satisfied are:
$ \require{begingroup}\begingroup
\def \a #1#2{ {\aleph} \!\!\: \left( {#1} , {#2} \right) }
\def \b #1{ {#1}\d{#1} }
\def \d #1{ {\large{{#1} \over 2019}} }
\def \e { {\!\;\varepsilon} }
\def \f #1{ \left\lfloor {#1} \right\rfloor }
\def \l { \\[.3ex] }
\def \x { {\-\b1} }
\def \xd { {\-\!\:\d{2020}} }
\def \xp { {\big( \x \big)} }
\def \. #1{ {\,{#1}\,} }
\def \- { {\scriptsize \raise.25ex -} }
\def \+ { {\scriptsize \phantom +} }
\def \={ \kern-.3em & \kern-.3em = \kern-.3em & \kern-.3em }
$
\begin{array}{rcccccc} n \= \+4035 \\[1ex] x \= \xd \= \-\b{1} \\[2ex] \a{x}{n} \= \-2019 \= {\Large{ 2020 \over \-\,\LARGE{2020\over2019}~ }} \= \Large{2020 \over \Large \raise.3ex x} \end{array}
This solution uses the recurrence relation
$\a{x}{i{+}1} = \f{x\,\a{x}{i}}$
observed in the definition
$ \a{x}{i} = \underbrace{\f{x\,\f{x\,\f{...\f{x}}}}}
_{\large \f{~i~\sf levels~}} $.
Here is how $x\,\a{x}{n} = 2020$ is reached:
\begin{matrix} \a{x}{1} \= \f{x } \= \f{\x } \= \-2 \l \a{x}{2} \= \f{x\,\a{x}{1}} \= \f{\+\b{2}} \= \+2 \l \a{x}{3} \= \f{x\,\a{x}{2}} \= \f{\-\b{2}} \= \-3 \l \a{x}{4} \= \f{x\,\a{x}{3}} \= \f{\+\b{3} } \= \+3 \\[-.3ex] &\vdots& &\vdots& &\vdots& \\ \a{x}{4033} \= \f{x\,\a{x}{4032}} \= \f{\-\b{2017}} \= \-2018 \l \a{x}{4034} \= \f{x\,\a{x}{4033}} \= \f{\+\b{2018}} \= \+2018 \l \a{x}{4035} \= \f{x\,\a{x}{4034}} \= \f{\-\b{2018}} \= \-2019 \\[2ex] \hline \raise1ex\strut \boldsymbol{x\,\a{x}{n}} \= x\,\a{x}{4035} \= \xp(\-2019) \= \boldsymbol{2020} \end{matrix}
For this solution the goal is taken to approach
$~ \a{x}{i} = \large{2020 \over \large \raise.3ex x} ~$
as gradually as possible.
This suggests examining values of $x$
that border between progressing and getting stuck
along the recurrence relation $\a{x}{i{+}1} = \f{x\,\a{x}{i}}$.
It seems obvious that $\a{x}{i}$ should not overshoot 2020 for $i \.< n$
and that, for minimal progress, $x$ should be as close as possible to 0.
The puzzle statement’s example of $\a{\-{2020\over979}}{9}$
opens the way for $x \.< 0$
but it is easier to get a feel for the puzzle with $x \.> 0$.
What is the smallest positive value of $x$ that does not get stuck?
It is $x \.= 2$, as demonstrated in comparison to $1 \.\le x \.< 2$.
\begin{array}{rclcrcl}
\a{2}{1} \= \f{2} && \a{x}{1} \= \f{x} \kern1em\textsf{for$~~1\le x<2$} \\
\= 2 && \= 1 \\[1.5ex]
\a{2}{2} \= \f{2\f{2}} && \a{x}{2} \= \f{x\f{x}} \l
\= 4 && \= \f{(x)(1)} \\
\small \textsf{(doubles from}~\rlap{\textsf{$\a{2}{1}$ to $\a{2}{2}$)}}
&& &\kern3em& \= 1 \\
&& && & \small\llap {\textsf{(stuck at}}~\rlap{\textsf{$\a{x}{1}$)}}
\end{array}
For this smallest positive candidate of $x \. = 2$,
$~ \a{2}{i} $ grows exponentially to $\a{2}{11} = 2048$,
which is too much,
meaning that $n{=}10$ would be the largest possibility for $n$
if $x$ is some difficult-to-pin-down number near 2.
What, then, is the smallest (closest to zero)
negative value of $x$ that does not get stuck?
It is $x = \-1{-}\e$ with an infinitesimally positive $\e$,
as demonstrated in comparison to $x \.= \-1$.
\begin{array}{rclcrcl}
\a{\-1{-}\e}{1} \= \f{\-1{-}\e} && \a{\-1}{1} \= \f{\-1} \\
\= \-2 && \= \-1 \\[1.5ex]
\a{\-1{-}\e}{3} \= \f{(\-1{-}\e)\f{(\-1{-}\e)\f{\-1{-}\e}}}
&& \a{\-1}{3} \= \f{(\-1)\f{(\-1)\f{\-1}}} \l
\= \f{(\-1{-}\e)\f{(\-1{-}\e)(-2)}} && \= \f{(\-1)(1)} \l
\= \f{(\-1{-}\e)\f{2{+}2\e}} &\kern1em& \= \-1 \l
\= \f{(\-1{-}\e)(2)} &&
& \small\llap{\textsf{(stuck at}}~\rlap{\textsf{$\a{\-1}{1}$)}} \l
\= \f{\-2{-}2\e} \\
\= \-3 \\
\small \textsf {($\,$increments by}
~\rlap{\textsf{$\-1$ from $\, \a{\-1{-}\e}{1} \,$ to $\, \a{\-1{-}\e}{3} \,$)}}
\end{array}
Pursuing this candidate of $x = \-1{-}\e$ works as well as could be hoped!
Not only is the progression of $\a{x}{i}$ linear rather than exponential
but it also grows at only half the rate of $i$,
as laid out for the solution’s $x$ near the top of this answer.
All that remains is to choose an $x$ near −1
so that $x \, \a{x}{n} = 2020$.
Although $~ \a{\-1{-}\e}{4037} = \-2020 ~$
looks promising, it is too good to be true because
$~ (\-1{-}\e)\,\a{\-1{-}\e}{4037} = 2020{+}2020\e > 2020 ~$
overshoots the target.
Thus, using $n \.= 4035$ and working from
$~ \a{\-1{-}\e}{4035} = \-2019 ~$
means solving for $~ x = \-1{-}h ~$ in
$~ (\-1{-}h)\,\a{\-1{-}h}{4035} = 2019{+}2019h = 2020 \,$.
And there it is, $\, h \.= \d{1} \,$ so $\, x \.= \-\b{1} \,$.
$\endgroup$
