FINAL ROUND

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 $61-n$ numbers in the initial set $M=\{n,n+1,\dots ,60\}$, we have $k\le \frac{61-n}{3}$. 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

($S(n)=\frac{(n+60)(61-n)}{2}$, and the sum of all numbers of all subsets is less than or equal to $$2\cdot (60+59+58+\cdots +(61-k))=2\cdot \frac{(60+61-k)\cdot k}{2}=(121-k)\cdot k\le$$ $$\le \left(121-\frac{61-n}{3}\right)\cdot \frac{61-n}{3}=\frac{(302+n)(61-n)}{9}.$$ (The last inequality holds because $k\le (61-n)/3\le 121/2$ and the function $(121-x)x$ increases for $x\le 121/2$.) Therefore, the following condition is necessary to exist the desired partition: $$S(n)\le \frac{(302+n)(61-n)}{9}\iff \frac{(n+60)(61-n)}{2}\le \frac{(302+n)(61-n)}{9}\iff$$ $$9(n+60)\le 2(302+n)\iff 7n\le 64,$$ i.e., $n\le 9^{1/7}$, and since $n$ is an integer number, we have $n\le 9$. The sum $S(n)=\frac{(n+60)(61-n)}{2}$ is even, so either $n=4m$ or $n=4m+1$, i.e., $n\in \{1,4,5,8,9\}$. If $n=9$, then $S(9)=\frac{(9+60)(61-9)}{2}=69\cdot 26=1794$. The number $k$ of the subsets is less than or equal to $k\le \frac{61-9}{3}=17\frac{1}{3}$, i.e., $k\le 17$. But for these $k$ the sum of all numbers in the subsets is less than or equal to $(121-k)\cdot k=104\cdot 17=1768<S(9)$, a contradiction. Similarly, if $n=8$, then $S(8)=1804$, $k=[53:3]=17$, and $1768<S(8)$, a contradiction.

60	59	58	57	56	55	54	53	52
43	41	39	37	35	33	31	29	27
17	18	19	20	21	22	23	24	25

51	50	49	48	47	46	45	44	26
42	40	38	36	34	32	30	28	8,6
9	10	11	12	13	14	15	16	7,5

60	59	58	57	56	55	54	53	52
43	41	39	37	35	33	31	29	27
17	18	19	20	21	22	23	24	25
51	50	49	48	47	46	45	44	28
42	40	38	36	34	32	30	26	16
9	10	11	12	13	14	15	8,6,4	7,5

$n=4$

If we add the subset $\{1,2,3\}$ to the given partition for $n=4$ we obtain the desired partition for $n=1$.

The desired partitions exist for $n=1,4,5$. The following tables give the examples of the desired partition for $n=5$ and $n=4$:

60	59	58	57	56	55	54	53	52
43	41	39	37	35	33	31	29	27
17	18	19	20	21	22	23	24	25

51	50	49	48	47	46	45	44	26
42	40	38	36	34	32	30	28	8,6
9	10	11	12	13	14	15	16	7,5

$n=5$

If we add the subset $\{1,2,3\}$ to the given partition for $n=4$ we obtain the desired partition for $n=1$.