Four, there are four numbers $1, a, b, c$ satisfying $1<a<b<c$ and their pairwise sums are all different. These 6 sums, when arranged in ascending order, form an arithmetic sequence with a total sum of 201. Find the values of $a, b, c$.

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Four, there are four numbers $1, a, b, c$ satisfying $1<a<b<c$ and their pairwise sums are all different. These 6 sums, when arranged in ascending order, form an arithmetic sequence with a total sum of 201. Find the values of $a, b, c$.
Obviously, $1+a<1+b<1+c<a+c<b+c$,
$$
\begin{array}{r}
1+a<1+b<a+b<a+c<b+c . \\
\text { Also, } 1+a+b+c=\frac{201}{3}=67, \\
a+b+c=66
\end{array}
$$

If $a+b<1+c$, then the common difference is
$$
b-a=a-1=1+c-a-b
$$

Thus,
$$
b=2 a-1, c=2 b-1=4 a-3
$$

Substituting into (1) yields $a=10, b=19, c=37$. In this case, the 6 sums are
$$
11,20,29,38,47,56
$$

which meet the requirements.

If $1+c<a+b$, then the common difference is
$$
b-a=c-b=a+b-c-1
$$

Thus,
$$
c=2 a-1, b=\frac{1}{2}(c+a)=\frac{1}{2}(3 a-1) .
$$

Substituting into (1) yields $a=15, b=22, c=29$. In this case, the 6 sums are
$$
16,23,30,37,44,51
$$

which also meet the requirements.