I divide 1989 into the sum of 10 positive integers, to maximize their product.
1. When 1989 is decomposed into: $199+199+\cdots+199+198$, the corresponding product is $199^{9} \times$ 198. Below is the proof that it is the largest. Since the number of decompositions is finite, the maximum value must exist, so we can assume the decomposition $\left(x_{1}, x_{2}, \cdots, x_{10}\right)$, where $x_{1}+x_{2}+\cdots+x_{10}=1989$, is one of the decompositions with the largest product $x_{1} x_{2} \cdots x_{10}$. If any of $x_{1}, x_{2}, \cdots, x_{10}$ is less than 198, let it be $x_{1}$, then there must be one greater than 198, let it be $x_{10}$. Let $x_{1}^{\prime}=x_{1}+\left(x_{10}-198\right), x_{10}^{\prime}=198$, then we get a new decomposition $\left(x_{1}^{\prime}, x_{2}, \cdots, x_{9}, x_{10}^{\prime}\right)$, with the corresponding product $x^{\prime}{ }_{1} x_{2} \cdots x_{9} x_{10}^{\prime}$. But $x_{1}^{\prime} x_{10}^{\prime}-x_{1} x_{10}=\left[x_{1}+\left(x_{10}-198\right)\right] \cdot 198-x_{1} x_{10}=198 x_{1}+198 x_{10}-198^{2}-$ $x_{1} x_{10}=\left(198-x_{1}\right)\left(x_{10}-198\right)>0$, so $x_{1}^{\prime} x_{2} \cdots x_{9} x_{10}^{\prime}>x_{1} x_{2} \cdots x_{10}$, which is a contradiction. Therefore, $x_{1}, x_{2}, \cdots, x_{10}$ are all not less than 198. If any of $x_{1}, x_{2}, \cdots, x_{10}$ is greater than 199, let it be $x_{1}$, then there must be one less than 199, let it be $x_{10}$. Since $x_{10}$ is not less than 198, so $x_{10}=$ 198. Let $x_{1}^{\prime}=x_{1}-1, x_{10}^{\prime}=x_{10}+1$, then we get a new decomposition $\left(x_{1}^{\prime}, x_{2}, \cdots, x_{9}\right.$, $\left.x_{10}^{\prime}\right)$, with the corresponding product $x_{1}^{\prime} x_{2} \cdots x_{9} x_{10}^{\prime}$. But $x_{1}^{\prime} x_{10}^{\prime}-x_{1} x_{10}=\left(x_{1}-1\right)\left(x_{10}+1\right)-$ $x_{1} x_{10}=x_{1} x_{10}+x_{1}-x_{10}-1-x_{1} x_{10}=x_{1}-x_{10}-1=x_{1}-198-1=x_{1}-$ $199>0$, so $x_{1}^{\prime} x_{2} \cdots x_{9} x_{10}^{\prime}>x_{1} x_{2} \cdots x_{10}$, which is a contradiction. Therefore, $x_{1}, x_{2}, \cdots, x_{10}$ can only be 199 or 198, thus the decomposition: $199+199+\cdots+199+198$ corresponds to the product $199^{9} \times 198$ being the largest.