<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] There are two ways of interpreting this question. The first is "what is the maximum possible product for a (finite) sequence of natural numbers whose sum is 20". The second is "Fix the natural number $n$. What is the maximum possible product for a sequence of $n$ natural numbers whose sum is $n$".
I assumed the first interpretation was correct. In which case the answer is as follows:
$2n\le n^2$ for $n\ge 2$ and $2n+1<n(n+1)$ for $n\ge 2$. So we do not need to use any number bigger than 3 to get the maximum value. Clearly there is no benefit in using 1s. We have $3^2>2^3$, so it is better to use 3s rather than 2s. The largest number available is six 3s which takes us up to 18. Hence the maximum is $2\cdot3^6=1458$.
However, others claim that the second interpretation is correct. So I will deal briefly with that. For any natural number $k$ we have $k(k+2)<(k+1)^2$, so the maximising sequence cannot have two numbers with a difference of two or more. Clearly we must have $n$ in the range $1-20$ of it is not possible to satisfy the condition that the numbers sum to 20. Call $M_n$ the maximum product for $n$ numbers.
So for $n=1$ we have $M_1=20$. We have $M_2=100$ with solution $10,10$. $M_3=294$ with solution $6,7,7$. $M_4=625$ with solution $5,5,5,5$. $M_5=1024$ with solution $4,4,4,4,4$. $M_6=1296$ with solution $3,3,3,3,4,4$. $M_7=1458$ with solution $3,3,3,3,3,3,2$. $M_8=1296$ with solution $3,3,3,3,2,2,2,2$. $M_9=1152$ with solution $3,3,2,2,2,2,2,2,2$. $M_{10}=1024$ with solution all 2s. $M_{11}=512,M_{12}=256,M_{13}=128,M_{14}=64,M_{15}=32$, $M_{16}=16,M_{17}=8,M_{18}=4$ ,$M_{19}=2,M_{20}=1$ with a mix of 1s and 2s.
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