9. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University).
Answer: 1770.

Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:

$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1755 \Leftrightarrow 1000 a+100 b+10 c+d=a+b+c+d+1755 \Leftrightarrow \\
\Leftrightarrow 999 a+99 b+9 c=1755 \Leftrightarrow 111 a+11 b+c=195
\end{gathered}
$$

From the last equation, we conclude that the only possible value for the digit $a$ is 1. Next, we get the equation $11 b+c=84$, which means the digit $b$ can only be 7. Finally, we need to find the value of $c=84-11 \cdot 7=7$. Therefore, the desired number is 1770, as it is the smallest among all numbers of the form $\overline{177 d}$.