(2) Furthermore, for any $1 \leqslant i, j \leqslant 5$ there is $\left|x_{i}-x_{j}\right| \leqslant 2$, when $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ take what values, $S$ reaches the minimum value. (2006 National High School Mathematics League Question)
(2) When $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$ and $| x_{i}-x_{j} | \leqslant 2$, only the following three cases satisfy the requirements:
( i ) $402,402,402,400,400$;
( ii ) $402,402,401,401,400$;
( iii) $402,401,401,401,401$.
The latter two cases are obtained by adjusting $x^{\prime}_{i}=x_{i}-1, x^{\prime}_{j}=x_{j}+1$ from the first set of cases. According to the proof in part (1), each adjustment increases the sum $S=\sum_{1 \leq i<j \leqslant 5} x_{i} x_{j}$, so the minimum value of $S$ is achieved when $x_{1}=x_{2}=x_{3}=402, x_{4}=x_{5}=400$.