14. (IMO-30 Shortlist) Let $m$ be a positive odd number, find the smallest natural number $n$ such that $2^{1989}$ divides $m^{n}-1$.

 untranslated text remains the same as requested.
14. Let $n$ be represented as $n=2^{*} q$, where $q$ is an odd number, then
$$
\begin{aligned}
m^{n}-1 & =m^{2^{5} q}-1=\left(m^{2 s}\right)^{q}-1 \\
& =\left(m^{2^{5}}-1\right)\left[\left(m^{2^{5}}\right)^{q-1}+\left(m^{2^{5}}\right)^{q-2}+\cdots+m^{2^{x}}+1\right] \\
& =\left(m^{2^{5}}-1\right) A, \text { where } A \equiv 1(\bmod 2) .
\end{aligned}
$$

Thus, $2^{1989}\left|\left(m^{n}-1\right) \Leftrightarrow 2^{1989}\right|\left(m^{2^{x}}-1\right)$.
Therefore, we can set $n=2^{s}$, in which case there are two scenarios:
(1) If $m \equiv 1(\bmod 4)$, then the binary representation of $m$ is $m=\underbrace{1 \cdots \cdots}_{k \uparrow 10001}$, i.e., $k$ is the largest integer such that $m \equiv 1\left(\bmod 2^{k}\right)$. Thus, $m^{2}-1=(m+1)(m-1)$ is divisible by $2^{k+1}$ but not by $2^{k+2}$. Suppose $m^{2^{s}}-1$ is divisible by $2^{k+1}$ but not by $2^{k+t+1}$, then $m^{2^{i+1}}-1=\left(m^{2^{i}}+1\right)\left(m^{2^{i}}-1\right)$ is divisible by $2^{k+t+1}$ but not by $2^{k+1+2}$.

Therefore, for all natural numbers $s, 2^{s+k} \mid\left(m^{2^{s}}-1\right), 2^{s+k+1} \times\left(m^{2^{s}}-1\right)$. The largest integer, similarly, can be proven for all natural numbers $s, 2^{s+k} \mid\left(m^{2^{x}}-1\right), 2^{s+k+1} \times\left(m^{2^{x}}-1\right)$.
Thus, from $2^{1989} \mid\left(m^{2^{x}}-1\right) \Rightarrow 1989 \leqslant s+t \Rightarrow s \geqslant 1989-k$ (where $k$ is defined as above).
Therefore, when $k \leqslant 1989$, the smallest exponent $n=2^{1989-k}$, and when $k>1989$, $n=2^{0}=1$.