7. Represent the number 2015 as the sum of some number of natural numbers so that their product is the largest.

Answer. $2015=3+3+\ldots+3+2$ (671 threes and one two).
Solution If the sought representation contains at least one addend greater than 4, then the product is not maximal. Indeed, if $n \geq 5$ is replaced by the sum $2+(n-2)$, then the product $2 \cdot(\mathrm{n}-2)$ is greater than $\mathrm{n}$. Also, note that if the representation contains an addend 4, it can be replaced by $2+2$, and the product will remain unchanged.

Thus, the maximum product is obtained when the number is broken down into a sum of twos and threes (it is not difficult to understand that the replacements $2 \rightarrow 1+1$ and $3 \rightarrow 2+1, 2+2 \rightarrow 3+1$ only decrease the product). The replacement $2+2+2 \rightarrow 3+3$ does not change the sum but increases the product, so the number of twos should be as few as possible.

From this, we get the decomposition $2015=671 \cdot 3+2$, i.e., the number 2015 should be represented as a sum of 672 addends, one of which is equal to two, and the rest are equal to three.