1.1. Determine whether there exists a positive integer $n$ such that the sum of the digits of $n^{2}$ is 2002 .
1.1 Take $n=10^{222}-3$. Then
$$
\left(10^{222}-3\right)^{2}=10^{444}-6 \cdot 10^{222}+9=\underbrace{9 \cdots 9}_{221 \text { nines }} 40 \cdots 09 .
$$

The sum of the digits of $n^{2}=222 \times 9+4=2002$.