Reciprocals of Odd Numbers adding to 1

Theorem
$105$ is the smallest positive integer $n$ such that $1$ can be expressed as the sum of reciprocals of distinct odd integers such that none are less than $\dfrac 1 n$:


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {33} + \dfrac 1 {35} + \dfrac 1 {45} + \dfrac 1 {55} + \dfrac 1 {77} + \dfrac 1 {105}$

There are $5$ ways of expressing $1$ as the sum of reciprocals of only $9$ distinct odd integers, but then the least term is less than $\dfrac 1 {105}$.

The $9$-term solutions are as follows:


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {35} + \dfrac 1 {45} + \dfrac 1 {231}$


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {231} + \dfrac 1 {315}$


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {33} + \dfrac 1 {45} + \dfrac 1 {385}$


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {165} + \dfrac 1 {693}$


 * $1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {135} + \dfrac 1 {10 \, 395}$