$1 \cdot 15$ Given 5 distinct positive numbers can be divided into two groups such that the sums of the numbers in each group are equal, how many different ways are there to divide these numbers into such groups?



The above text translated into English, keeping the original text's line breaks and format, is as follows:
$1 \cdot 15$ Given 5 distinct positive numbers can be divided into two groups such that the sums of the numbers in each group are equal, how many different ways are there to divide these numbers into such groups?
[Solution] (1) If the number of elements in the two groups with equal sums are 1 and 4, respectively, i.e., the largest number among the 5 numbers equals the sum of the other 4 numbers, then there is obviously only 1 way to divide them.
(2) Suppose the number of elements in the two groups with equal sums are 2 and 3, respectively: $\left\{m_{1}, m_{2}\right\}$, $\left\{m_{3}, m_{4}, m_{5}\right\}$, and $m_{1}+m_{2}=m_{3}+m_{4}+m_{5}$. From (1), we know that if there is another valid grouping, it must still be a 2,3 grouping. Assume without loss of generality that it is $m_{1}+m_{5}=m_{2}+m_{3}+m_{4}$, but this leads to $m_{5}=m_{2}$, which is a contradiction.
Therefore, there is only 1 way to group the numbers that satisfies the requirements of the problem.