Let E denote the set of all natural numbers n such that 3 < n < 100...

$\{1,2 \ldots \ldots \ldots . . n\}$ This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3 , $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ is divisible by 3 . So n will be of the form $3 \lambda, 3 \lambda+2$ or for convenience we can take $\mathrm{n}=6 \lambda, 6 \lambda+2,6 \lambda+3,6 \lambda+5$ If $\mathrm{n}=6 \lambda$ then we can group numbers in bundles of 6 . In each bundle we can select numbers like $1,2,3,4,5,6(16,25,34)$ If $n=6 \lambda+2$ then we can club last bundle of 8 numbers rest can be partitioned and those 8 numbers can be done $1,2,3,4,5,6,7,8(1236,48)$ If $\mathrm{n}=6 \lambda+3$, we can club last nine numbers and rest can be partitioned $1,2,3,4,5,6,7,8,9 (12345,69,78)$ If $n=6 \lambda+5$ we can take last five numbers. Rest can be partitioned $1,2,3,4,5(14,23,5)$ Hence we can select any number of form $6 \lambda(16), 6 \lambda+2(16), 6 \lambda+3(16), 6 \lambda+5(16)$ so total 64 numbers.