32. For $n \geqslant 4$, let $k_{1}, k_{2}, \cdots, k_{n}$ be a permutation of $1,2, \cdots, n$, and satisfy $0 < a_{k_{1}} < a_{k_{2}} < \cdots < a_{k_{n}} \leqslant m$. Thus, $a_{k_{i}} - a_{k_{i-1}} \geqslant \frac{1}{k_{i} + k_{i-1}} (i=2,3, \cdots, n)$. By the Cauchy-Schwarz inequality, $m \geqslant a_{k_{n}} - a_{k_{1}} = \sum_{i=2}^{n} \left(a_{k_{i}} - a_{k_{i-1}}\right) \geqslant \sum_{i=2}^{n} \frac{1}{k_{i} + k_{i-1}} \geqslant \frac{(n-1)^{2}}{\sum_{i=2}^{n} \left(k_{i} + k_{i-1}\right)} = \frac{(n-1)^{2}}{n(n+1) - k_{1} - k_{n}} \geqslant \frac{(n-1)^{2}}{n^{2} + n - 3} = 1 - \frac{3n - 4}{n^{2} + n - 3}$. Note that this inequality holds for any $n \geqslant 4$, hence $m \geqslant 1$.