From $a_1$ to $a_{200}$ , then back again

$\normalsize\begin{aligned}&\dfrac {a_{1}}{a_{2}} + \dfrac {a_{2}}{a_{3}} + \dfrac {a_{3}}{a_{4}} + ... + \dfrac {a_{199}}{a_{200}} + \dfrac {a_{200}}{a_{1}} = 200\hspace{5mm}\small \color{#3D99F6}\text{Given}\\\\ &\text{Since, }{a_{1}},a_{2},a_{3}\cdots a_{200}\text{ are positive reals,}\\\\ &\text{Applying AM-GM we get, }\\\\\end{aligned}$

$\normalsize\begin{aligned}\dfrac{\dfrac{a_{1}}{a_{2}} + \dfrac {a_{2}}{a_{3}} + \dfrac {a_{3}}{a_{4}} + ... + \dfrac {a_{199}}{a_{200}}+ \dfrac {a_{200}}{a_{1}}}{200}&\geq\sqrt[200]{\dfrac {a_{1}}{a_{2}}\cdot\dfrac {a_{2}}{a_{3}}\cdot\dfrac {a_{3}}{a_{4}}\cdots\dfrac {a_{199}}{a_{200}}\cdot\dfrac {a_{200}}{a_{1}}}\\\\ \dfrac{\dfrac{a_{1}}{a_{2}} + \dfrac {a_{2}}{a_{3}} + \dfrac {a_{3}}{a_{4}} + ... + \dfrac {a_{199}}{a_{200}}+ \dfrac {a_{200}}{a_{1}}}{200}&\geq1\hspace{15mm}\small\color{#3D99F6}\text{As }\dfrac {a_{1}}{a_{2}}\cdot\dfrac {a_{2}}{a_{3}}\cdot\dfrac {a_{3}}{a_{4}}\cdots\dfrac {a_{199}}{a_{200}}\cdot\dfrac {a_{200}}{a_{1}}=1\\\\ \implies\dfrac{a_{1}}{a_{2}} + \dfrac {a_{2}}{a_{3}} + \dfrac {a_{3}}{a_{4}} + ... + \dfrac {a_{199}}{a_{200}}+ \dfrac {a_{200}}{a_{1}}&\geq200\\\\\end{aligned}$

$\normalsize\begin{aligned}&\text{The equality only occurs at,} \dfrac{a_{1}}{a_{2}} =\dfrac {a_{2}}{a_{3}} =\dfrac {a_{3}}{a_{4}} = ... = \dfrac {a_{199}}{a_{200}}= \dfrac {a_{200}}{a_{1}}\\\\ &\text{But since }{a_{1}}<a_{2}<a_{3}<\cdots <a_{200}\text{ we have,}\\\\ &\dfrac{a_{1}}{a_{2}} , \dfrac {a_{2}}{a_{3}} , \dfrac {a_{3}}{a_{4}} , ... , \dfrac {a_{199}}{a_{200}}<1 \text{ and } \dfrac {a_{200}}{a_{1}}>1\\\\ \implies&\dfrac{a_{1}}{a_{2}} + \dfrac {a_{2}}{a_{3}} + \dfrac {a_{3}}{a_{4}} + ... + \dfrac {a_{199}}{a_{200}}+ \dfrac {a_{200}}{a_{1}}>200\\\\ &\text{Making the number of solutions } \color{#EC7300}\boxed{\color{#333333}0}\end{aligned}$

Suppose $(a_1, a_2, a_3, \ldots, a_{200})$ is such a solution.

Since $0 < a_1 < a_2 < a_{200}$ , we have $\frac{a_1}{a_2} < 1 < \frac{a_{200}}{a_1}$ , so in particular $\frac{a_1}{a_2} \neq \frac{a_{200}}{a_1}$ .

Therefore, by the [strict] AM-GM inequality, $1 = \sqrt[200]{1} = \sqrt[200]{\frac{a_1}{a_2}\times\frac{a_2}{a_3}\times\cdots\times \frac{a_{199}}{a_{200}}\times \frac{a_{200}}{a_1}} < \frac{1}{200}\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots + \frac{a_{199}}{a_{200}} + \frac{a_{200}}{a_1}\right) = \frac{1}{200}(200) = 1.$

That is, $1 < 1$ , which of course is absurd, so there are no such solutions.