6. Find the smallest positive integer $n$ or prove that no such $n$ exists, satisfying the following property: there exist infinitely many different $n$-tuples of positive rational numbers $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, such that $\sum_{i=1}^{n} a_{i}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}}$ are both integers.
6. $n=3$.

If $n=1$, then $a_{1} \in \mathbf{Z}_{+}, \frac{1}{a_{1}} \in \mathbf{Z}_{+}$. Therefore, $a_{1}=1$, which does not satisfy the condition.
If $n=2$, and there exist positive rational numbers $x, y$, such that
$$
\begin{array}{l}
x+y \in \mathbf{Z}_{+}, \frac{1}{x}+\frac{1}{y} \in \mathbf{Z}_{+}. \\
\text {Let } x=\frac{a}{b}, y=\frac{c}{d}, \text { where } a, b, c, d \in \mathbf{Z}_{+}, \\
\text {and } (a, b)=(c, d)=1. \text { Then } \\
x+y \in \mathbf{Z}_{+} \Leftrightarrow b d \mid(a d+b c), \\
\left.\frac{1}{x}+\frac{1}{y} \in \mathbf{Z}_{+} \Leftrightarrow a c \right\rvert\,(a d+b c).
\end{array}
$$

From equation (1), we get
$d \mid (a d+b c) \Leftrightarrow d \mid b c \Leftrightarrow d \mid b$.
From equation (1) again, we get
$b \mid (a d+b c) \Leftrightarrow b \mid a d \Leftrightarrow b \mid d$.
Thus, $b=d$.
Similarly, from equation (2), we get $a=c$.
Therefore, $x=\frac{a}{b}=\frac{c}{d}=y$.
Thus, the problem reduces to finding all $x \in \mathbf{Q}_{+}$ such that $2 x \in \mathbf{Z}_{+}, \frac{2}{x} \in \mathbf{Z}_{+}$.
Let $n=2 x \in \mathbf{Z}_{+}$. Then
$\frac{2}{x} \in \mathbf{Z}_{+} \Leftrightarrow \frac{4}{n} \in \mathbf{Z}_{+} \Leftrightarrow n=1,2$ or 4.
Thus, $(x, y)$ has only a finite number of solutions $\left(\frac{1}{2}, \frac{1}{2}\right), (1,1)$, or $(2,2)$, which does not satisfy the condition.

If $n=3$, we will prove: there exist infinitely many triples of positive rational numbers $(x, y, z)$ such that
$$
x+y+z \in \mathbf{Z}, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \in \mathbf{Z}.
$$

Let $x+y+z=1$, and
$$
(x, y, z)=\left(\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}\right),
$$

where $a, b, c \in \mathbf{Z}_{+}$.
$$
\begin{array}{l}
\text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \\
=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c} \in \mathbf{Z} \\
\Leftrightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \in \mathbf{Z}.
\end{array}
$$

Fix $a=1$, we need to prove there exist infinitely many pairs of positive integers $(b, c)$ such that
$$
\begin{array}{l}
\frac{1}{b}+\frac{1}{c}+\frac{c}{b}+\frac{b}{c}=3 \\
\Leftrightarrow b^{2}+c^{2}-3 b c+b+c=0.
\end{array}
$$

Clearly, $b=2, c=3$ satisfies equation (3).
Assume $c \geqslant b$, and satisfies equation (3).
Rewrite equation (3) as a quadratic equation in $b$ for a fixed $c$:
$$
b^{2}-(3 c-1) b+\left(c^{2}+c\right)=0.
$$

Let the other root of equation (4) be $b_{0}$.
By Vieta's formulas, we have
$$
b+b_{0}=3 c-1, b b_{0}=c^{2}+c.
$$

Since $b, c \in \mathbf{Z}$, we know $b_{0} \in \mathbf{Z}$;
Since $b, c \in \mathbf{Z}_{+}$, we know
$b_{0} \in \mathbf{Z}_{+}$, and $b_{0}=\frac{c^{2}+c}{b} \geqslant \frac{c^{2}+c}{c}>c$.
Thus, from the solution $(b, c)$, we can obtain another solution $(c, b_{0})$ of equation (3), and $b_{0}>c$.

Repeating this method, we can obtain infinitely many different solutions of equation (3), the first few of which are
$$
(2,3),(3,6),(6,14),(14,35),(35, \\
90),(90,234),(234,611),(611,1598), (1598,4182), \ldots
$$
In conclusion, the minimum value of $n$ is 3.