Find the maximum product

Let's consider a partition of natural number $N>1$ into sum: $N = { a }_{ 1 }+...+{ a }_{ n }$ . For any ${ a }_{ i }$ if it equals 1, we can raise the value of the product by merging it with any other number, since: $1\cdot { a }_{ i }<\left( { a }_{ i }+1 \right)$ . Now let's take a look at numbers that are greater than 4, ${ a }_{ i }>4$ :

${ a }_{ i }>4\\ { a }_{ i }>4.5\\ 2\cdot { a }_{ i }>9\\ 3\cdot { a }_{ i }>9+{ a }_{ i }\\ 3\cdot \left( { a }_{ i }-3 \right) >{ a }_{ i }$ , while $3+\left( { a }_{ i }-3 \right) ={ a }_{ i }$ . That means that we can split every such number into 3's, 4's and 2's, and the product will only increase.

Then we can split every 4 into two 2's without any difference, and thus we've got a product of 2's and 3's. Since $2\cdot 2\cdot 2<3\cdot 3$ , we can transform most of the 2's into 3's with two, one or none left depending on $N mod 3$ . Because we can do this with any partition, this partition in particular gives us the maximum product.

So when $N=14$ , we have one 2 and four 3's, meaning the final answer is $\boxed { 162 }$