15. Express 2006 as the sum of 5 positive integers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. Let $S=\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}$, find:
(1) For what values of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, does $S$ achieve its maximum value;
15. (1) First, the set of such $S$ values is bounded, so there must exist a maximum and a minimum. If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, and $S=\sum_{1 \leqslant i < j \leqslant 5} x_{i} x_{j}$,

Rewrite $S$ as
$$S=x_{1} x_{2}+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}+x_{5}\right)+x_{3} x_{4}+x_{3} x_{5}+x_{4} x_{5}$$

At the same time, we have
$$S^{\prime}=x^{\prime}{ }_{1} x^{\prime}{ }_{2}+\left(x_{1}^{\prime}+x^{\prime}{ }_{2}\right)\left(x_{3}+x_{4}+x_{5}\right)+x_{3} x_{4}+x_{3} x_{5}+x_{4} x_{5}$$

Thus, $S^{\prime}-S=x^{\prime}{ }_{1} x^{\prime}{ }_{2}-x_{1} x_{2}>0$, which contradicts the assumption that $S$ achieves its maximum value at $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. Therefore, it must be that $\left|x_{i}-x_{j}\right| \leqslant 1(1 \leqslant i, j \leqslant 5)$. Hence, the maximum value is achieved when $x_{1}=402, x_{2}=x_{3}=x_{4}=$ $x_{5}=401$.