Five, (15 points) Find the smallest positive integer $n$ such that $2^{2005}$ । $\left(161^{n}-1\right)$.

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Five, (15 points) Find the smallest positive integer $n$ such that $2^{2005} \mid \left(161^{n}-1\right)$.
Let $n=2^{s} q, s \in \mathbf{N}, 2$ does not divide $q$. Then $161^{n}-1=161^{2^{s} q}-1=\left(161^{2^{s}}-1\right) A$.
It is easy to see that 2 does not divide $A$.
Given $2^{2005} \mid\left(161^{n}-1\right)$, we get $2^{2005} \mid\left(161^{2^{3}}-1\right)$.
Thus, we can set $n=2^{x}$.
Since $161=2^{5} \times 5+1$, we will prove that $2^{x+5} \|\left(161^{2^{3}}-1\right)$.
When $s=1$, the above equation clearly holds.
Assume that when $s=k$, we have $2^{k+5} \|\left(161^{2^{k}}-1\right)$.
Then when $s=k+1$,
$161^{2^{k+1}}-1=\left(161^{2^{k}}\right)^{2}-1$
$=\left(161^{2^{k}}-1\right)\left(161^{2^{k}}+1\right)$.
It is easy to see that $2 \|\left(161^{2^{k}}+1\right)$ and $2^{k+5} \|\left(161^{2^{k}}-1\right)$, thus $2^{k+6} \|\left(161^{2^{k+1}}-1\right)$.
Therefore, $k+5 \geqslant 2005$, i.e., $k \geqslant 2000$.
Thus, the smallest positive integer $n$ such that $2^{2005} \mid \left(161^{n}-1\right)$ is $2^{2000}$.