5. (55th Moscow Mathematical Olympiad, 1992) Among the odd natural numbers $n$ less than 10000, are there more odd numbers $n$ for which the number formed by the last 4 digits of $n^9$ is greater than $n$, or more odd numbers $n$ for which the number formed by the last 4 digits of $n^9$ is less than $n$?
5. The same amount.

Divide the odd natural number $n$ less than 10000 into several groups, each containing two numbers $\left(n_{1}, n_{2}\right)$, and $n_{1}+n_{2}=10000$.
Since $n_{1}+n_{2} \mid n_{1}^{9}+n_{2}^{9}$,
it follows that $10000 \mid n_{1}^{9}+n_{2}^{9}$.
Let the last 4 digits of $n_{1}^{9}$ form the number $m_{1}$, and the last 4 digits of $n_{2}^{9}$ form the number $m_{2}$. Then from (1) we get $m_{1}+m_{2}=10000$.
This implies $n_{1}>m_{1}$ is equivalent to $m_{2}>n_{2}$.
Therefore, the number of four-digit numbers mentioned in the problem is the same.