For $n=1, a_{1} \in \mathbb{Z}_{>0}$ and $\frac{1}{a_{1}} \in \mathbb{Z}_{>0}$ if and only if $a_{1}=1$. Next we show that (i) There are finitely many $(x, y) \in \mathbb{Q}_{>0}^{2}$ satisfying $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$. Write $x=\frac{a}{b}$ and $y=\frac{c}{d}$ with $a, b, c, d \in \mathbb{Z}_{>0}$ and $\operatorname{gcd}(a, b)=\operatorname{gcd}(c, d)=1$. Then $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$ is equivalent to the two divisibility conditions
$$b d \mid a d+b c \quad(1) \quad \text { and } \quad a c \mid a d+b c$$
Condition (1) implies that $d|a d+b c \Longleftrightarrow d| b c \Longleftrightarrow d \mid b$ since $\operatorname{gcd}(c, d)=1$. Still from (1) we get $b|a d+b c \Longleftrightarrow b| a d \Longleftrightarrow b \mid d$ since $\operatorname{gcd}(a, b)=1$. From $b \mid d$ and $d \mid b$ we have $b=d$. An analogous reasoning with condition (2) shows that $a=c$. Hence $x=\frac{a}{b}=\frac{c}{d}=y$, i.e., the problem amounts to finding all $x \in \mathbb{Q}_{>0}$ such that $2 x \in \mathbb{Z}_{>0}$ and $\frac{2}{x} \in \mathbb{Z}_{>0}$. Letting $n=2 x \in \mathbb{Z}_{>0}$, we have that $\frac{2}{x} \in \mathbb{Z}_{>0} \Longleftrightarrow \frac{4}{n} \in \mathbb{Z}_{>0} \Longleftrightarrow n=1,2$ or 4, and there are finitely many solutions, namely $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right),(1,1)$ or $(2,2)$. (ii) There are infinitely many triples $(x, y, z) \in \mathbb{Q}_{>0}^{2}$ such that $x+y+z \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \in \mathbb{Z}$. We will look for triples such that $x+y+z=1$, so we may write them in the form
$$(x, y, z)=\left(\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}\right) \quad \text { with } a, b, c \in \mathbb{Z}_{>0}$$
We want these to satisfy
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c} \in \mathbb{Z} \Longleftrightarrow \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c} \in \mathbb{Z}$$
Fixing $a=1$, it suffices to find infinitely many pairs $(b, c) \in \mathbb{Z}_{>0}^{2}$ such that
$$\frac{1}{b}+\frac{1}{c}+\frac{c}{b}+\frac{b}{c}=3 \Longleftrightarrow b^{2}+c^{2}-3 b c+b+c=0$$
To show that equation (*) has infinitely many solutions, we use Vieta jumping (also known as root flipping): starting with $b=2, c=3$, the following algorithm generates infinitely many solutions. Let $c \geqslant b$, and view (*) as a quadratic equation in $b$ for $c$ fixed:
$$b^{2}-(3 c-1) \cdot b+\left(c^{2}+c\right)=0$$
Then there exists another root $b_{0} \in \mathbb{Z}$ of ( $\left.* *\right)$ which satisfies $b+b_{0}=3 c-1$ and $b \cdot b_{0}=c^{2}+c$. Since $c \geqslant b$ by assumption,
$$b_{0}=\frac{c^{2}+c}{b} \geqslant \frac{c^{2}+c}{c}>c$$
Hence from the solution $(b, c)$ we obtain another one $\left(c, b_{0}\right)$ with $b_{0}>c$, and we can then "jump" again, this time with $c$ as the "variable" in the quadratic (*). This algorithm will generate an infinite sequence of distinct solutions, whose first terms are $(2,3),(3,6),(6,14),(14,35),(35,90),(90,234),(234,611),(611,1598),(1598,4182), \ldots$
Comment. Although not needed for solving this problem, we may also explicitly solve the recursion given by the Vieta jumping. Define the sequence $\left(x_{n}\right)$ as follows:
$$x_{0}=2, \quad x_{1}=3 \quad \text { and } \quad x_{n+2}=3 x_{n+1}-x_{n}-1 \text { for } n \geqslant 0$$
Then the triple
$$(x, y, z)=\left(\frac{1}{1+x_{n}+x_{n+1}}, \frac{x_{n}}{1+x_{n}+x_{n+1}}, \frac{x_{n+1}}{1+x_{n}+x_{n+1}}\right)$$
satisfies the problem conditions for all $n \in \mathbb{N}$. It is easy to show that $x_{n}=F_{2 n+1}+1$, where $F_{n}$ denotes the $n$-th term of the Fibonacci sequence ( $F_{0}=0, F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$ for $n \geqslant 0$ ).