. Decompose 2008 into a sum of natural numbers whose product is maximal.
. Let $2008=a_{1}+\cdots+a_{k}$ be a partition that maximizes the product. We start by noting that if one of the $a_{i}$ is greater than 5, then by replacing it with 3 and $a_{i}-3$, we strictly increase the product. Thus, all the $a_{i}$ are 2, 3, or 4. We also note that replacing a 4 with two 2s does not change the product, so without loss of generality, we can reduce the case to where all the $a_{i}$ are 2 or 3. Finally, since $2^{3}<3^{2}$, we cannot have more than two 2s. In conclusion, the partition that maximizes the product is obtained by taking 668 times the number 3 and two times the number 2.