3. (25 points) On a circle, there are $n$ different positive integers $a_{1}$, $a_{2}, \cdots, a_{n}$ placed in a clockwise direction. If for any number $b$ among the ten positive integers $1, 2, \cdots, 10$, there exists a positive integer $i$ such that $a_{i}=b$ or $a_{i}+a_{i+1}=b$, with the convention that $a_{n+1}=a_{1}$, find the minimum value of the positive integer $n$.
3. From the conditions, we know that the $2n$ numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ should include the ten positive integers $1,2, \cdots, 10$. Therefore, $2 n \geqslant 10 \Rightarrow n \geqslant 5$.

When $n=5$, the ten numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n}+a_{1}$ are all distinct and take values from $1 \sim 10$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{n}+\left(a_{1}+a_{2}\right)+ \\
\left(a_{2}+a_{3}\right)+\cdots+\left(a_{n}+a_{1}\right) \\
=1+2+\cdots+10 \\
\Rightarrow 3\left(a_{1}+a_{2}+\cdots+a_{n}\right)=55,
\end{array}
$$

which is a contradiction.
Thus, when $n=5$, there do not exist five positive integers $a_{1}, a_{2}, \cdots, a_{5}$ that satisfy the conditions.

When $n=6$, construct $a_{1}=1, a_{2}=10, a_{3}=2, a_{4}=6, a_{5}=3, a_{6}=4$, and verify that these six numbers satisfy the conditions.
In conclusion, the minimum value of the positive integer $n$ is 6.