3. (10 points) $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ are natural numbers satisfying $0<a_{1}<a_{2}<a_{3} \cdots<a_{n}$, and $\frac{13}{14}=\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}+\cdots$ $+\frac{1}{a_{n}}$, then the minimum value of $n$ is . $\qquad$
【Analysis】To make $n$ the smallest, it means to minimize the number of terms, so each term should be as small as possible. $0\frac{1}{3}$,
Let $a_{2}=3$, then $\frac{1}{\mathrm{a}_{3}}+\cdots+\frac{1}{\mathrm{a}_{\mathrm{n}}}=\frac{3}{7}-\frac{1}{3}=\frac{2}{21}>\frac{1}{11}$,
Let $a_{3}=11$, then $\frac{1}{\mathrm{a}_{4}}+\cdots+\frac{1}{\mathrm{a}_{\mathrm{n}}}=\frac{2}{21}-\frac{1}{11}=\frac{1}{231}$, so $a_{4}=231$.
Therefore, the minimum value of $n$ is 4.
【Solution】Solution: Let $\frac{13}{14}\frac{1}{3}$,
Let $a_{2}=3$, then $\frac{1}{\mathrm{a}_{3}}+\cdots+\frac{1}{\mathrm{a}_{\mathrm{n}}}=\frac{3}{7}-\frac{1}{3}=\frac{2}{21}>\frac{1}{11}$,
Let $a_{3}=11$, then $\frac{1}{\mathrm{a}_{4}}+\cdots+\frac{1}{\mathrm{a}_{\mathrm{n}}}=\frac{2}{21}-\frac{1}{11}=\frac{1}{231}$, so $a_{4}=231$.
Therefore, the minimum value of $n$ is 4.