Solution:

We write all fractions of the form $b / a$, where $a$ and $b$ are relatively prime, and $0 \leq b \leq a \leq n$, in ascending order. For instance, for $n=5$, this is the sequence
$$\frac{0}{1}, \frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{1}{1}$$

This sequence is known as the Farey sequence.
Now, if we look at the the sequence of the denominators of the fractions, we see that $k$ appears $\varphi(k)$ times when $k>1$, although 1 appears twice. Thus, there are $N+1$ elements in the Farey sequence. Let the Farey sequence be

$$\frac{b_{1}}{a_{1}}, \frac{b_{2}}{a_{2}}, \ldots, \frac{b_{N+1}}{a_{N+1}}$$

Now, $a_{N+1}=1$, so the sequence $a_{1}, a_{2}, \ldots, a_{N}$ contains $\varphi(k)$ instances of $k$ for every $1 \leq k \leq n$. We claim that this sequence also satisfies
$$\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{N} a_{1}}=1 .$$

Since $a_{1}=a_{N+1}=1$, we have
$$\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{N} a_{1}}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{N} a_{N+1}} .$$

Now, it will suffice to show that $\frac{1}{a_{i} a_{i+1}}=\frac{b_{i+1}}{a_{i+1}}-\frac{b_{i}}{a_{i}}$. Once we have shown this, the above sum will telescope to $\frac{b_{N+1}}{a_{N+1}}-\frac{b_{1}}{a_{1}}=1-0=1$.
To see why $\frac{1}{a_{i} a_{i+1}}=\frac{b_{i+1}}{a_{i+1}}-\frac{b_{i}}{a_{i}}$ holds, we note that this is equivalent to $1=b_{i+1} a_{i}-b_{i} a_{i+1}$. We can prove this fact geometrically: consider the triangle in the plane with vertices $(0,0),\left(a_{i}, b_{i}\right)$, and
$\left(a_{i+1}, b_{i+1}\right)$. This triangle contains these three boundary points, but it contains no other boundary or interior points since $a_{i}$ and $a_{i+1}$ are relatively prime to $b_{i}$ and $b_{i+1}$, respectively, and since no other fraction with denominator at most $n$ lies between $\frac{b_{i}}{a_{i}}$ and $\frac{b_{i+1}}{a_{i+1}}$. Thus, by Pick's theorem, this triangle has area $1 / 2$. But the area of the triangle can also be computed as the cross product $\frac{1}{2}\left(b_{i+1} a_{i}-b_{i} a_{i+1}\right)$; hence $b_{i+1} a_{i}-b_{i} a_{i+1}=1$ and we are done.