$2 \cdot 121$ Let $M$ be a subset of the set $\{1,2, \cdots, 25\}$, such that the sums of the numbers in any two disjoint subsets of $M$ are not equal. Find the maximum possible sum of all the numbers in $M$.
［Solution］Set
$$
M_{0}=\{25,24,23,21,18,12\}
$$

satisfies the requirements of the problem, where the sum of the 6 numbers is 123, so the required maximum value is no less than 123.
Let $A=\{1,2, \cdots, 25\}$. We first prove that any 7-element subset $S$ of $A$ cannot satisfy the requirements of the problem. If not, the number of all non-empty subsets of $S$ with elements no greater than 4 is
$$
C_{7}^{1}+C_{7}^{2}+C_{7}^{3}+C_{7}^{4}=98,
$$

and the sum of all numbers in each of these subsets is a natural number no greater than $22+23+24+25=94$. By the pigeonhole principle, there must be two subsets with the same sum of numbers, which is a contradiction.

Secondly, we prove that any 5-element subset of $\{25,24, \cdots, 19\}$ cannot satisfy the requirements of the problem. In fact, the 10 differences obtained by taking the difference of any two numbers in a 5-element subset are all natural numbers no greater than 6. Therefore, either there are 3 equal differences, or there are two equal differences and the difference is no less than 4. Both scenarios lead to two equal differences, and the 4 numbers forming these two differences are all different, thus forming two disjoint 2-element subsets with equal sums.

The above argument shows that the set $M$ that meets the requirements is a 6-element set, and the sum of the 5 larger elements is no greater than $25+24+23+21+18=111$. To make the sum greater than 123, the smallest number must be increased. However, if the first 5 numbers remain unchanged, the smallest number being greater than 12 does not meet the requirements, so when the smallest number is increased, one of the first 5 numbers must be decreased, and if only one number is reduced by 1, it also does not meet the requirements. If the total reduction of the first 5 numbers is no less than 2, to make the sum greater than 123, the smallest number should be no less than 15.

Finally, we prove that any 6-element subset of $\{25,24, \cdots, 15\}$ does not satisfy the requirements of the problem. If not, the 6 numbers can form 15 differences, each of which is a natural number no greater than 10. From the previous proof, there cannot be 3 equal differences, nor can there be two equal differences with a value no less than 6, so the 15 differences can only be 2 each of $1,2,3,4,5$ and 1 each of $6,7,8,9,10$.

In this way, the 6-element subset $T$ must include 25 and 15, and one of 24 and 16, let's assume it is 24. Thus, $16 \notin T$. Since there are two differences of 5, $20 \in T$. Besides $25-24=1$, there must be another difference of 1, and the two must have a common element, so $23 \in T$. Therefore, $22,21 \notin T$. Since $24-20=4$, and there is another difference of 4, $16 \in T$, which is a contradiction, completing the entire proof.
The required maximum value is 123.