Three, prove using the proof by contradiction.
Suppose the set is divided into two subsets $A$ and $B$, in both $A$ and $B$, the equation $x_{1} + x_{2} + \cdots + x_{9} = x_{10}$ has no solution.
If $1 \in A$, to make the equation unsolvable, since
$$x_{1} + x_{2} + \cdots + x_{9} \geqslant 9 \text {, }$$

then $9 \in B$. Otherwise, if $9 \in A$, then $x_{1} = x_{2} = \cdots = x_{9} = 1, x_{10} = 9$ is a solution to the equation.

Therefore, $81 \in A$, otherwise, if $81 \in B$, by taking $x_{1} = x_{3} = \cdots = x_{9} = 9 \in B, x_{10} = 81 \in B$, the equation has a solution.

At this point, if $10 \in A$, then by $81 = 10 \times 8 + 1$, taking $x_{1} = x_{2} = \cdots = x_{8} = 10, x_{9} = 1, x_{10} = 81$, since $10 \in A, 1 \in A$, $81 \in A$, the equation has a solution in $A$, leading to a contradiction.

If $10 \in B$, since $89 = 10 \times 8 + 9$. Then taking $x_{1} = x_{2} = \cdots = x_{8} = 10 \in B, x_{9} = 9 \in B$, 89 must be in $A$. At this time, by
$$89 = 1 \times 8 + 81 \text {, }$$

taking $x_{1} = x_{3} = \cdots = x_{8} = 1 \in A, x_{9} = 81 \in A, x_{10} = 89 \in A$. The equation has a solution in $A$, leading to a contradiction.

Thus, for any partition of $M = \{1, 2, \cdots, 89\}$ into $A$ and $B$, the equation $x_{1} + x_{2} + \cdots + x_{9} = x_{10}$ has a solution in one of the subsets.

Note: $M = \{1, 2, \cdots, 89\}$ is the smallest set such that the equation $x_{1} + x_{2} + \cdots + x_{9} = x_{10}$ has a solution in any partition of $M = \{1, 2, \cdots, n\}$.

If $M' = \{1, 2, \cdots, 88\}$, then a partition can be constructed such that the equation has no solution in both subsets $A$ and $B$, for this, we can set
$$\begin{array}{l} A = \{1, 2, 3, \cdots, 8, 81, 82, \cdots, 89\}, \\ B = \{9, 10, 11, \cdots, 79, 80\} . \end{array}$$
(Provided by Tian Qiao Wang Lianxiao)