Adding unit fractions

$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{n} < 1$

$\frac{41}{42} + \frac{1}{n} < 1$

$\frac{1}{n} < \frac{1}{42}$

$n > 42$

The next positive integer such that $n > 42$ is $n = \boxed{43}$ .

We require that $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{n} \lt 1 = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} \Longrightarrow \dfrac{1}{7} + \dfrac{1}{n} \lt \dfrac{1}{6} \Longrightarrow \dfrac{1}{n} \lt \dfrac{1}{6} - \dfrac{1}{7} = \dfrac{1}{42} \Longrightarrow n \gt 42$ .

The least positive integer $n$ is thus $\boxed{43}$ .

For smallest value we don't need to work much if we observe that $\begin{aligned} \dfrac{1}{2}+\dfrac{1}{2}& = 1\implies \dfrac{1}{2}+\dfrac{1}{\text{l.c.m}(2,1)+1} <1 \\\dfrac{1}{2} +\dfrac{1}{3} + \dfrac{1}{6}& =1\implies \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{\text{l.c.m(2,3)}} <1\end{aligned}$ Therefore, $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7} = \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{\text{l.c.m} (2,3)+1} <1$ On the same manner $\dfrac{1}{2} +\dfrac{1}{3}+\dfrac{1}{7} +\dfrac{1}{n} =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7} +\dfrac{1}{\text{l.cm}(2,3,7)+1 }< 1$ giving us $n= \text{l.cm}(2,3,7)+1=43$ as the smallest possible value. The next possible value will be $\text{l.c.m}\,(2,3,7,43)+1 = 1807$ and so on.

Note: $\text{l.c.m}(2,3,7,43,1807)+1=3263443$ is the prime.