17 Positive integer $n$ cannot be divided by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
Solve for $n=1$, $\left|2^{1}-3^{1}\right|=1$; for $n=5$, $\left|2^{2}-3^{2}\right|=5$; for $n=7$, $\left|2^{1}-3^{2}\right|=7$; for $n=11$, $\left|2^{4}-3^{3}\right|=11$; for $n=13$, $\left|2^{4}-3^{1}\right|=13$; for $n=17$, $\left|2^{6}-3^{4}\right|=17$; for $n=19$, $\mid 2^{3}-$ $3^{3} \mid=19$; for $n=23$, $\left|2^{5}-3^{2}\right|=23$; for $n=25$, $\left|2^{1}-3^{3}\right|=$ $25$; for $n=29$, $\left|2^{5}-3^{1}\right|=29$; for $n=31$, $\left|2^{5}-3^{0}\right|=31$;

Next, we prove that $n=35$ satisfies the requirement using proof by contradiction. If not, there exist non-negative integers $a, b$ such that $\left|2^{a}-3^{b}\right|=35$.
(1) If $2^{a}-3^{b}=35$, clearly $a \neq 0,1,2$, so $a \geqslant 3$. Modulo 8, we get $-3^{b} \equiv 3(\bmod 8)$, which means $3^{b} \equiv 5(\bmod 8)$. However, $3^{k} \equiv 1,3(\bmod 8)$, which is impossible.
(2) If $3^{b}-2^{a}=35$, it is easy to see that $b \neq 0,1$. Modulo 9, we get $2^{a} \equiv 1(\bmod 9)$. And $\left\{2^{k}(\bmod 9)\right\}: 2,4,8,7,5,1,2,4, \cdots$, that is:
$$
\begin{array}{l}
2^{6 k} \equiv 1(\bmod 9), 2^{6 k+1} \equiv 2(\bmod 9), 2^{6 k+2} \equiv 4(\bmod 9), \\
2^{6 k+3} \equiv 8(\bmod 9), 2^{6 k+4} \equiv 7(\bmod 9), 2^{6 k+5} \equiv 5(\bmod 9),
\end{array}
$$

Thus, $a=6 k$, where $k$ is a non-negative integer, so $3^{b}-8^{2 k}=35$.
Modulo 7, we get $3^{b} \equiv 1(\bmod 7)$. And $\left\{3^{k}(\bmod 7)\right\}: 3,2,6,4,5,1, 3,2, \cdots$, so $b=6 k^{\prime}$, where $k^{\prime}$ is a positive integer, so
$$
3^{6 k^{\prime}}-2^{6 k}=35,
$$

which means $\left(3^{3 k^{\prime}}-2^{3 k}\right)\left(3^{3 k^{\prime}}+2^{3 k}\right)=35$,

Thus, $3^{3 k^{\prime}}=18$ or 6, which is impossible.
In conclusion, $n_{\min }=35$.