We note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into 2 subsets with sum 39: $\{1,2,3,10,11,12\}$ and $\{4,5,6,7,8,9\}$. The next largest possible sum is 77 (the entire set except 1). If $k \geq 2$ subsets each had equal sum, then they would have to be 7 subsets with sum 11 each or 11 subsets with sum 7 each. However, the subset containing 12 will have sum greater than 11; hence there is no way to partition the subset $\{2, \ldots, 12\}$ into equal subsets.