6.31 Find a set of five distinct natural numbers such that any two of them are coprime, and the sum of any several of them (more than 1) is a composite number.
(21st All-Soviet Union Mathematical Olympiad, 1987)
［Solution］We consider the general case:
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ numbers satisfying
$$
a_{i}=i \cdot n!+1, \quad i=1,2, \cdots, n .
$$

Then any two of these $n$ numbers are coprime.
Otherwise, if $\left(a_{i}, a_{j}\right)=d>1$, then from
$$
\begin{array}{l}
a_{i}=i \cdot n!+1=p d, \\
a_{j}=j \cdot n!+1=q d .
\end{array}
$$

we get
$$
(i-j) n!=(p-q) d \text {. }
$$

Thus $d \mid(i-j) n$!
From (1), we know that $d$ is not a divisor of $2,3, \cdots, n$, and since $|i-j|<n$, it follows that $d=1$, meaning $a_{i}$ and $a_{j}$ are coprime.
Moreover, the sum of any $k$ numbers among $a_{1}, a_{2}, \cdots, a_{n}$ can be divided by $k$.
Therefore, $a_{i}=i \cdot n!+1, i=1,2, \cdots, n$ are $n$ numbers that satisfy the given conditions.
Specifically, when $n=5$, these five numbers are
$$
121,241,361,481,601 \text {. }
$$