(1) Find $f(2006)$;
(2) If $f(n) \geqslant 30$, find the minimum value of $n$.

For a positive integer $n$, let $f(n)$ be the sum of the digits of $n^2$.
(1) Since $2006^{2}=4024036$, we have
$$
f(2006)=19.
$$
(2) Since $f(n) \geqslant 30$, then $n^{2} \geqslant 3999$.
Thus, $n>63$.
$$
\begin{array}{l}
\text{And } f(64)=19, f(65)=13, f(66)=18, \\
f(67)=25, f(68)=16, f(69)=18, \\
f(70)=13, f(71)=10, f(72)=18, \\
f(73)=19, f(74)=22, f(75)=18, \\
f(76)=25, f(77)=25, f(78)=18, \\
f(79)=13, f(80)=10, f(81)=18, \\
f(82)=19, f(83)=31,
\end{array}
$$

Therefore, the smallest positive integer $n$ such that $f(n) \geqslant 30$ is 83.
Note: Using a CASIO $\mathrm{fx}-82 \mathrm{MS}$ calculator, you can press the following keys:
$$
63=\text{ Ans }^{2}: \sqrt{\text{ (Ans) }}+1 \text{, }
$$

Then, repeatedly press the $“=$" key to get the values of $64^{2}, 65^{2}, 66^{2}, \cdots$.
Using a TI-83Plus or T1-92Plus calculator, you can press the following keys:
$$
63 \rightarrow \mathrm{N} \text{ ENTER } 1+\mathrm{N} \rightarrow \mathrm{N}:\left\{\mathrm{N}, \mathrm{N}^{2}\right\},
$$

Then, repeatedly press the “ENTER” key to get
$$
\{64,4096\},\{65,4225\}, \cdots .
$$