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 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:
$\color{blue}{1+11}\color{brown}{+2+10}\color{blue}{+4+8}\color{brown}{+5+7}\color{blue}{+3+6}\color{brown}{+9}$
$=\color{blue}{1+3+4+6+8+11}\color{brown}{+2+5+7+9+10}$
