Hint. $$\frac{1}{n} +\frac{1}{n} = \frac{1}{n} + \frac{1}{n+1}+\frac{1}{n(n+1)}.$$
For example, say $N=3$. Then we can write:
$$\begin{align*}
1 &= \frac{1}{4}+\frac{1}{4} + \frac{1}{4}+\frac{1}{4}\\
&= \frac{1}{4} + \frac{1}{5}+\frac{1}{20} + \frac{1}{5}+\frac{1}{20}+\frac{1}{5}+\frac{1}{20}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{20} + \frac{1}{6}+\frac{1}{30} + \frac{1}{21}+\frac{1}{420} + \frac{1}{6}+\frac{1}{30} + \frac{1}{21}+\frac{1}{420}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{20}+\frac{1}{6}+\frac{1}{30}+\frac{1}{21}+\frac{1}{420} + \frac{1}{7}+\frac{1}{42} + \frac{1}{31}\\
&\qquad\mathop{+}\frac{1}{930} + \frac{1}{22}+\frac{1}{462} + \frac{1}{421}+\frac{1}{176820}\\
&= \frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{30}+\frac{1}{31}+\frac{1}{42}\\
&\qquad\mathop{+}\frac{1}{420}+\frac{1}{421}+\frac{1}{462}+\frac{1}{930}+\frac{1}{176820}.
\end{align*}$$