Solution:

Assume the contrary. Then there exists a positive integer multiple of $u_{m}$ such that the sum of its digits is less than $m$ and let $t$ be the smallest number with this property. Since $t > 10^{m}$, the number $t$ can be written as $t = 10^{m} a + b$, where $0 < b < 10^{m}$.

We have $t = 10^{m} a + b = (10^{m} - 1) a + a + b$. Since $u_{m}$ divides both $t$ and $10^{m} - 1 = 9 u_{m}$, we conclude that $u_{m}$ divides $a + b$. But the sum of the digits of $a + b$ does not exceed the sum of the digits of $t$ and $a + b < t$, a contradiction.