The partition problem is a well-known NP-complete problem. We focus on its optimization version. This problem has been extensively studied from various perspectives over time. We propose two partially ordered sets (posets) corresponding to the partition problem and present a novel methodology for solving the problem. The first poset is order-isomorphic to a well-known poset whose structure is related to solutions of the subset sum problem, while the second is a subposet of the first and plays a crucial role in this paper. We first show several properties of the two posets, such as size, height, and width (the size of the largest antichain, i.e., the largest set of pairwise incomparable elements). Both widths are the same and \(\mathrm {\Theta }(2^n / n^{3/2})\) for n congruent to 0 or 3 modulo 4, which indicates the hardness of the partition problem. We then prove that the initial candidate solutions are the elements of the second poset, whose size is \(2^{n} - 2 \left( {\begin{array}{c}n\\ \lfloor n/2 \rfloor \end{array}}\right) \) . Since a partition corresponds to two elements of the poset, the number of initial candidate partitions is half of that, i.e., \(2^{n-1} - \left( {\begin{array}{c}n\\ \lfloor n/2 \rfloor \end{array}}\right) \) . We finally prove that candidate solutions can be reduced based on the partial order. In particular, we give several polynomially solvable cases by considering the minimal and maximal elements of the second poset. Our approach offers a useful tool for structural analysis of the partition problem.