For what values of $n$ can {1, 2, . . . , n} be partitioned into three subsets with equal sums? For what values of n can {1, 2, . . . , n} be partitioned into three subsets
with equal sums?
I noticed that somehow the sum from 1 to n hast to be a multiple of 3 and the common sum among these 3 subset is this sum divided by 3, but it's still not a convincing argument. How do you prove there exists 3 subsets of operands such that their sum evaluate to this number?
My solution made from other solutions
Theorem: The given set can be partitioned into three subsets with equals sums whem $\sum_{i = 1}^{n}i  = \frac{1}{2} n(n+1) \equiv 0 \mod 3$ except when $n = 3$ , so $n \equiv 0,2 \mod 3$
Here are some cases which will be important to prove our theorem
 n = 5 $\{1,4\},\{2,3\},\{5\}$
 n = 6 $\{1,6\},\{2,5\},\{3,4\}$
 n = 8 $\{8,4\},\{7,5\},\{1,3,6\}$
 n =9 $\{9,6\}, \{8,7\}, \{1,2,3,4,5\}$
Also note that if $n \equiv  0,2\mod 3$ then $n \equiv 0,2,3,5 \mod 6 $ and ${5,6,8,9}$ represents this modularity, so we can start creating subsets with equal sums with the last 6 elements of the n elements. Once we are done with this 6 elements, we continue with the next 6 until we get to 5,6,8 or 9 and then we apply the base case.
 A: If $n=6k$, make the following subsets:
$$1,4,\cdots,3k-2,3k+3,3k+6,\cdots,6k$$
$$2,5,\cdots,3k-1,3k+2,3k+5,\cdots,6k-1$$
$$3,6,\cdots,3k,3k+1,3k+4,\cdots,6k-2$$
If $n=6k+r$, where $r=5,8,9$, reduce to the previous case by first partitioning the set $\{1,2,\cdots,r\}$ and then partitioning the $6k$-element set $\{r+1,r+2,\cdots,n\}$ the same as above, except that each set is “translated upwards” by $r$ (all elements increased by $r$).
Example: partition $\{1,2,\cdots,11\}$: $r=5, k=1$ so we first partition $\{1,2,3,4,5\}$ into $\{1,4\},\{2,3\},\{5\}$, partition $\{1,2,3,4,5,6\}$ into $\{1,6\},\{2,5\},\{3,4\}$, “translate” the latter upwards by 5: $\{6,11\},\{7,10\},\{8,9\}$ and finally join them to get $\{1,4,6,11\},\{2,3,7,10\},\{5,8,9\}$.
The above construction works for any $n\ge 5$, $n\equiv 0$ or $n\equiv 2 \mod 3$. Cases $n=2$ and $n=3$ have no solutions, and neither have the cases $n\equiv 1 \mod 3$, because in those cases the total sum $\frac{n(n+1)}{2}$ is not divisible by $3$.
A: Clearly we need $n\equiv 0,2\bmod 3$ for the sum to be a multiple of $3$.
We need two disjoint sets with sum $s=\frac{n(n+1)}{6}$.
For the first set $A$ take $1,3,5,\dots$ until the next number would exceed $s$.
For the second set $B$ take $2,4,\dots $ until the next number would exceed $s$.
How many numbers have not been used?
the sum of the remaining numbers is at least $a=\frac{n(n+1)}{6}$ and so there are at least $\frac{n+1}{6}$ remaining numbers.
In other words the numbers $\underbrace{n,n-1,\dots}_a $ are unused.
We can use these numbers to fix $A$ and $B$.(more on this when Im up to it)
A: The sum of all the numbers from $1$ to $n$ is $\frac 12n(n+1)$.  As you say, we need this to be a multiple of $3$, which will be true when $n \equiv 0,2 \pmod 3$.  We can't do $n=2$ or $n=3$, which we can prove by inspection.  For $n=5$ there is $\{1,4\},\{2,3\},\{5\}$ and for $6$ we can do $\{1,6\},\{2,5\},\{3,4\}$.  For larger $n$ it should be "obvious" that you have so much flexibility that it will be possible.  To prove that, we can do $8$ and $9$ by hand, which I leave to you.  Then you can do blocks of $6$ down from the top $\{k+1,k+6\},\{k+2,k+5\},\{k+3,k+4\}$ until you have $5,6,8,9$ left and use the solution we have for that.  Now that you have groups of three sets with the same sum, take one out of each group to form one of the sets with sum $\frac 13 \cdot \frac 12n(n+1)$
