A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is essentially increasing if $f(s) \leq f(t)$ holds whenever $s \leq t$ are real numbers such that $f(s) \neq 0$ and $f(t) \neq 0$.

Find the smallest integer $k$ such that for any 2022 real numbers $x_{1}, x_{2}, \ldots, x_{2022}$, there exist $k$ essentially increasing functions $f_{1}, \ldots, f_{k}$ such that
$$
f_{1}(n)+f_{2}(n)+\cdots+f_{k}(n)=x_{n} \quad \text { for every } n=1,2, \ldots, 2022
$$
The answer is 11 and, more generally, if 2022 is replaced by $N$ then the answer is $\left\lfloor\log _{2} N\right\rfloor+1$.

Bound. Suppose for contradiction that $2^{k}-1>N$ and choose $x_{n}=-n$ for each $n=1, \ldots, N$. Now for each index $1 \leq n \leq N$, define
$$
S(n)=\left\{\text { indices } i \text { for which } f_{i}(n) \neq 0\right\} \subseteq\{1, \ldots, k\} \text {. }
$$

As each $S(n t)$ is nonempty, by pigeonhole, two $S(n)$ 's coincide, say $S(n)=S\left(n^{\prime}\right)$ for $nx_{n^{\prime}}$ in that case due to the essentially increasing condition.
₪ Construction. It suffices to do $N=2^{k}-1$. Rather than drown the reader in notation, we'll just illustrate an example of the (inductive) construction for $k=4$. Empty cells are zero.
\begin{tabular}{r|rrrr} 
& $f_{1}$ & $f_{2}$ & $f_{3}$ & $f_{4}$ \\
\hline$x_{1}=3$ & 3 & & & \\
\hline$x_{2}=1$ & 10 & -9 & & \\
$x_{3}=4$ & \multicolumn{4}{l}{} \\
\hline$x_{4}=1$ & 100 & 200 & -299 & \\
$x_{5}=5$ & & 200 & -195 & \\
$x_{6}=9$ & 100 & & -91 & \\
$x_{7}=2$ & & & 2 & \\
\hline$x_{8}=6$ & 1000 & 2000 & 4000 & -6994 \\
$x_{9}=5$ & & 2000 & 4000 & -5995 \\
$x_{10}=3$ & 1000 & & 4000 & -4997 \\
$x_{11}=5$ & & & 4000 & -3995 \\
$x_{12}=8$ & 1000 & 2000 & & -2992 \\
$x_{13}=9$ & & 2000 & & -1991 \\
$x_{14}=7$ & 1000 & & & -993 \\
$x_{15}=9$ & & & & 9
\end{tabular}

The general case is handled in the same way with powers of 10 replaced by powers of $B$, for a sufficiently large number $B$.