Sum & Square Sum

With $a + b + c = d + e + f$ , we are looking for the number partition into six different numbers. Taking $a, b, c, d, e, f \in S$ for some space set of natural numbers, if $S = \{1, 2, 3, 4, 5, 6\}$ , whose elements are as least as possible, no equal summation is possible because $1+2+3+4+5+6 = 21$ , which can't be halved.

Hence, $7$ must be involved in $S$ . Then if if $S = \{1, 2, 3, 4, 5, 6, 7\}$ , we will have $1+2+3+4+5+6+7 = 28$ . Thus, the summation will be less than $\dfrac{28}{2} = 14$ because only six numbers can be selected, and the excluded one must be even to follow the constraint.

To yield the least summation, we will attempt to examine the trials by excluding the most possible even integer. First, by taking $6$ out, we will reach $11 = 7+3+1 = 5+4+2$ as the only possible partition. However, $7^2 +3^2 +1^2 \neq 5^2+4^2+2^2$ . It's not our desired solution.

Then by taking $4$ out, we will reach $12 = 7+3+2 = 6+5+1$ as the only possible partition, and $7^2 +3^2 +2^2 = 6^2+5^2+1^2 = 62$ , which works perfectly.

Therefore, the least possible summation is $\boxed{62}$ .

1+25+36=4+9+49=62