14.3,29 ** Find a natural number $n$, such that each of $n, n+1, n+2, \cdots, n+20$ has a common divisor greater than 1 with $30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
Let $m-n+10$, then $n, n+1, \cdots, n+20$ are $m \pm 1, m \pm 2, \cdots, m \pm 10$. Since $11, 13, 2 \cdot 3 \cdot 5 \cdot 7$ are pairwise coprime, according to the Chinese Remainder Theorem, the system of congruences $m \equiv 0(\bmod 2 \cdot$ $3 \cdot 5 \cdot 7), m \equiv 1(\bmod 11), m \equiv -1(\bmod 13)$ has a positive integer solution $m$. At this time, $m, m \pm 2, m \pm 4, m \pm 6, m \pm 8, m \pm 10$ have a factor of $2; m \pm 3, m \pm 9$ have a factor of $3; m \pm 5$ have a factor of $5; m \pm 7$ have a factor of $7; m+1$ has a factor of $13; m-1$ has a factor of $11$. Therefore, $n=m-10$ is the desired solution. The smallest $n$ can be calculated as $9440$.