<s>[INST] If sum of n natural number is 20 then what is their max. product? If sum of n natural number is 20 then what is their max. product ?  
 [/INST] Let $s=20$ denote the given sum.
As almagest noted, $k(k+2)\lt(k+1)^2$ shows that the factors can differ by at most $1$. Thus there must be $j$ factors of $\left\lceil\frac sn\right\rceil$ and $n-j$ factors of $\left\lfloor\frac sn\right\rfloor$, with $j$ determined by
$$j\left\lceil\frac sn\right\rceil+(n-j)\left\lfloor\frac sn\right\rfloor=s$$
and thus
\begin{align}
j&=\frac{s-n\left\lfloor\frac sn\right\rfloor}{\left\lceil\frac sn\right\rceil-\left\lfloor\frac sn\right\rfloor}=s-n\left\lfloor\frac sn\right\rfloor=n\left(\frac sn-\left\lfloor\frac sn\right\rfloor\right)=n\left\{\frac sn\right\}=s\bmod n\;,
\end{align}
where $\{x\}$ is the fractional part of $x$. Thus, the solution has $s\bmod n$ factors of $\left\lceil\frac sn\right\rceil$ and $n-(s\bmod n)$ factors of $\left\lfloor\frac sn\right\rfloor$.
As shown here (and apparently also in almagest's answer, though I don't understand the argument there), if the intent was to maximise this over all $n$, the optimal case has only factors of $2$ and $3$, so it has either $\left\lfloor\frac sn\right\rfloor=2$, or $\frac sn=3$, and since $3^2\gt2^3$, among these cases it's the one with lowest $n$, which is $n=\left\lceil\frac s3\right\rceil$, and the product is
$$
3^{s\bmod\left\lceil\frac s3\right\rceil}2^{\left\lceil\frac s3\right\rceil-s\bmod\left\lceil\frac s3\right\rceil}=2^{\left\lceil\frac s3\right\rceil}\left(\frac32\right)^{s\bmod\left\lceil\frac s3\right\rceil}=3^{\left\lceil\frac s3\right\rceil}\left(\frac23\right)^{(-s)\bmod 3}\;.
$$
In the present case, for $s=20$, the optimal $n$ is $7$ and the maximal product is
$$3^{\left\lceil\frac{20}3\right\rceil}\left(\frac23\right)^{(-20)\bmod 3}=2\cdot3^6=1458\;.$$
</s>