20. When $n=2$, taking $m=1$ suffices. Suppose the proposition holds for $n(\geqslant 2)$, i.e., there exists $m \in \mathbf{N}^{*}$ such that $m^{3}+17=3^{n}(3 q+r)$, where $r \in\{1,2\}$ and $q \in \mathbf{N}$. In this case, $3 \nmid m$, hence $m^{2}=1(\bmod 3)$, and thus
$$3^{n} \cdot m^{2} \equiv 3^{n}\left(\bmod 3^{n+1}\right)$$

For any $s \in \mathbf{N}^{*}$, we have
$$\begin{aligned} \left(m+3^{n-1} \cdot s\right)^{3}+17 & =m^{3}+17+3^{n} \cdot m^{2} \cdot s+3^{2 n-1} \cdot m \cdot s^{2}+3^{3 n-3} \cdot s^{3} \\ & \equiv 3^{n}(3 q+r)+3^{n} \cdot s \equiv 3^{n}(r+s)\left(\bmod 3^{n+1}\right) \end{aligned}$$

Thus, taking $x_{1}=m+3^{n-1}(3-r), x_{2}=m+3^{n-1}(6-r)=x_{1}+3^{n}$, we have $3^{n+1} \mid\left(x_{1}^{3}+17\right)$ and $3^{n+1} \mid\left(x_{2}^{3}+17\right)$.

Furthermore, if $3^{n+2} \mid\left(x_{1}^{3}+17\right)$ and $3^{n+2} \mid\left(x_{2}^{3}+17\right)$, then
$$\begin{aligned} 0 & \equiv x_{2}^{3}+17=\left(x_{1}+3^{n}\right)^{3}+17=x_{1}^{3}+17+3^{n+1} x_{1}^{2}+3^{2 n+1} x_{1}+3^{3 n} \\ & \equiv x_{1}^{3}+17+3^{n+1} \equiv 3^{n+1}\left(\bmod 3^{n+2}\right) \end{aligned}$$

This is a contradiction (here we use $\left.x_{1}^{2} \equiv 1(\bmod 3)\right)$.
Therefore, the proposition holds for $n+1$ as well.