Answer: $n = 27^2 = 729$.
Since $n$ is odd, each of its divisors is odd. Clearly, the units digit of each divisor is odd.
Also, since the sum of the last digits of all the divisors is 33, which is an odd number, $n$ has an odd number of divisors. Hence, $n$ is a perfect square.
Since a prime number has only 2 divisors, and after squaring it has only 3 divisors, whose sum of last digits is less than 27, the three-digit numbers satisfying the condition can only be squares of composite numbers, of which there are four in total:
$$27^2, 25^2, 21^2, 15^2.$$
Checking them one by one, we find that $27^2 = 729$ is the desired answer.