Divide the numbers 1,2,…,n into two sets of equal sum. My approach:
If sum(1 to n) is divisible by 2, then it is possible to divide the set.
If n is even take 1st and last element and add to 1st set, then add 2nd and 2nd last to 2nd set...and so on, In then end to balance out both set I remove 1 from 1st set and add it to 2nd.
I don't know how to do it if n is odd.
For odd: eg: n=7
4
1 2 4 7 = 14
3
3 5 6 = 14

 A: This is essentially @Alexander Geldhof ideas.
We need to deal with the case when $n$ is odd and the sum $\sum_{i=1}^n i=\frac{n(n+1)}{2}$ being even. Note that these two conditions are equivalent to say that $n\equiv 3\mod 4$.
So, we need to prove the proposition for $n\equiv 3\mod 4$. Let's do it by induction.
Initial case $n=3$: Easily we can split $A=\{1,2\}$ $B=\{3\}$.
Inductive step $n\to n+4$:
Suppose we can split the set $\{1,\dots,n\} $ into two sets $A$,$B$ with the same sum. Then, we can split $\{1,\dots,n+4\}$ into the sets $A\cup \{n+1,n+4\}$ and $B\cup\{n+2,n+3\}$ note that these sets will remain with the same sum since $(n+1)+(n+4)=(n+2)+(n+3)$.
So the induction is complete.
Remark: In the case when $n$ is even and $\sum_{i=1}^ni$ is even, we conclude $n\equiv 0\mod 4$. The induction works as well with the initial case $n=4$, $A=\{1,4\}$, $B=\{2,3\}$.
A: First we know that there is no solution for $n=1$ nor for $n=2$. For $n=3$, we can form the sets $\{1, 2\}$ and $\{3\}$. For $n=4$, as you said, we can apply Gauss's method by forming the sets $\{1, 4\}$ and $\{2, 3\}$. If $n=5$, there is no solution, nor for $n=6$. However, for $n=7$, we can form the sets $\{1, 2, 4, 7\}$ and $\{3, 5, 6\}$. So now we have two possibilities: if $n-3$ is divisible by four, we can form the two sets $\{1, 2\}$ and $\{3\}$ and then apply Gauss's method starting with $4$: add $4$ and $n$ to one set, then $5$ and $n-1$, and so on. If $n$ is divisible by four, we can apply Gauss's method immediately. 
A: A triangular number is even if and only if its argument is $\in\{0,3\}\bmod 4$.
If the argument is $\equiv 0\bmod 4$, you can specify blocks from the "outside in".  For $n=8$ you put $1$ and $8$ into one block, then $2$ and $7$ into the second block, and alternate blocks until you hit the middle:
$1+2+3+4+5+6+7+8=\color{blue}{1+8}\color{brown}{+2+7}\color{blue}{+3+6}\color{brown}{+4+5}$
$=\color{blue}{1+3+6+8}\color{brown}{+2+4+5+7}$
For an argument $\equiv 3\bmod 4$, start by identifying the multiples of $m$ where the argument is $4m-1$.  There will be three such values $m,2m,3m$ such that $m+2m=3m$, so put $m+2m$ in one block and $3m$ in the other:
$1+2+3+4+5+6+7+8+9+10+11=1+2\color{blue}{+3}+4+5\color{blue}{+6}+7+8\color{brown}{+9}+10+11$
The remaining terms are partitioned outside in, the same way as for a multiple of $4$ terms described above:
$\color{blue}{1+11}\color{brown}{+2+10}\color{blue}{+4+8}\color{brown}{+5+7}\color{blue}{+3+6}\color{brown}{+9}$
$=\color{blue}{1+3+4+6+8+11}\color{brown}{+2+5+7+9+10}$
A: Note that the numbers for which the sum is even, and $n$ is odd, are exactly $n = \{3, 7, 11, \ldots \}$. It's clear how to divide the 3-set up. To divide the 7-set up, assign (4 and 7) to the first set and (5 and 6) to the second set. Continue this inductively.
