for 3 sets the sum of elements to be same sum upto n must be divisible by 3
or $\frac{n(n+1)}{2}$ should be divsible by 3
so n or n+ 1 must be divisible by 3
let us taken n divisible by 3
for n =3 we cannot divide into 3 equal groups
for 6 successive numbers say m+1 to m+ 6 we can devide into 3 groups $\{m+1,m+6\}, \{m+2,m+5\}.\{m+3,m+4\}$
so if n is multiple of 6 we can divide the numbers into groups as above
but if n is not divsible by 6 we can have a group of 9(1-9) and multiple groups of 6
the multiple groups of 6 can be broken into 3 groups and we can combine them into 3 groups
and 9 can be grouped as $\{4,9,2\},\{3,5,7\}, \{(1,6,8\}$ and take the elements and add to the above groups to get the result.
Let us take n + 1 divisible by 3 so n is of the form 3p + 2 where p >= 1
for p = 1 or n= 5 we can group as $\{4,1\},\{2,3\}, \{5\}$
for p = 2 or n= 8 we can group is $\{4,8\},\{1,5,6\}, \{(2,3,7\}$
for p > 2 we have number of term 5 + 6k or 8 + 6k and we can choose A,B,C accordingly as above
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