Let $m^\ast(I)$ the outer measure of $I$. Usually the outer measure is defined as $m^\ast(I):=\inf\left\{\sum_{k\in\Delta}\mathrm{length}(I_k):I\subseteq\bigcup_{k\in\Delta}I_k,\ \text{each $I_k$ is an open interval, $\Delta$ is a countable set}\right\}.$ The $J_n$ are a countable covering of $I$, so $\begin{align*} m^\ast(I) &\leq\sum_1^N \mathrm{length}(J_n)\notag \\ -m^\ast(I) &\geq -\sum_1^N \mathrm{length}(J_n)\notag \\ 1-m^\ast(I) &\geq 1-\sum_1^N \mathrm{length}(J_n)\tag{1}\label{eq} \end{align*}$ Suppose that $\sum_1^N \mathrm{length}(J_n)\lt 1$. From \ref{eq} follows $1-m^\ast(I)\gt 0.$ Remember that the outer measure of an interval is its length. By $\sigma$-subadditivity of the outer measure $\begin{align*} 1 &\leq m^\ast(I)+m^\ast(\Bbb Q)\\ 1-m^\ast(I) &\leq m^\ast(\Bbb Q), \end{align*}$ but $m^\ast(\Bbb Q)=0$ (you can see this as described here), so this says $1-m^\ast(I) \leq 0$ which contradicts \ref{eq}.
Therefore it must be $\sum_1^N \mathrm{length}(J_n)\geq 1$.
Notice that the same proof works fine with a countably infinite number of intervals in the cover of $I$.