Problem 1624

A number $a$ from a subset of the desired partition is called major if $a$ is equal to the sum of all other numbers of this subset. All numbers in each subset must be distinct, since one of them is major we see that there exist at least three numbers in each subset. Let $k$ be the number of the subsets of the desired partition. Since we have $25-n$ numbers in the initial set $M = \{n, n+1, \dots, 24\}$, we have $k \le \frac{25-n}{3}$. On the other hand, if $a$ is the major number of some subset, then the sum of the numbers of this subset is equal to $2a$. So the sum $S(n)$ of all numbers of the set $M$ must be even than or equal to
$$2 \cdot (24 + 23 + 22 + \dots + (25 - k)) = 2 \cdot \frac{(24 + 25 - k) \cdot k}{2} = (49 - k) \cdot k \le \\ \le \left(49 - \frac{25-n}{3}\right) \cdot \frac{25-n}{3} = \frac{(122+n)(25-n)}{9}.$$
(The last inequality holds because $k \le 25-n/3 \le 49/2$, and the function $(49-x)x$ increases for $x \le 49/2$.) Therefore, the following condition is necessary to exist the desired partition:
$$S(n) \le \frac{(122+n)(25-n)}{9} \iff \frac{(n+24)(25-n)}{2} \le \frac{(122+n)(25-n)}{9} \iff 9(n+24) \le 2(122+n) \iff n \le 4.$$
For $n=2$ and $n=3$ we see that both the sums
$$S(2) = \frac{(2+24)(25-2)}{2} = 13 \cdot 23, \quad S(3) = \frac{(3+24)(25-3)}{2} = 27 \cdot 11$$
are odd, so the desired partitions do not exist.