Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999?
(1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$

is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.

If the sum of the digits is 2 and it is an odd number, then the first and last digits must both be 1, with all the middle digits being 0. Let's assume it is $10^{a}+1$ $\left(a \in \mathbf{Z}_{+}\right)$.
$$
\begin{array}{l}
\text { Hence } 3 n^{2}+n+1=10^{a}+1 \\
\Rightarrow n(3 n+1)=10^{a} . \\
\text { But }(3 n+1, n)=(1, n)=1, \text { and } 3 n+1>n,
\end{array}
$$
Thus, $n=2^{a}, 3 n+1=5^{a}$.
So, $3 \times 2^{a}+1=5^{a}$.
But when $a>1$,
$$
\left(\frac{5}{2}\right)^{a}>3+1>3+\frac{1}{2^{a}},
$$

which is a contradiction.
Therefore, $a$ can only be 1.
Thus, $n=2$, and in this case, $3 n+1=7 \neq 5$, which is a contradiction.
Therefore, the sum of the digits must be at least 3.
When $n=8$, $3 n^{2}+n+1=201$, and the sum of the digits is exactly 3, so 3 is the minimum value.
$$
\begin{array}{l}
\text { (2) Let } n=10^{m}-1 \text {. Then } \\
3 n^{2}+n+1=3 \times 10^{2 m}-6 \times 10^{m}+3+10^{m} \\
=2 \underbrace{99 \cdots 95}_{m-1 \uparrow} \underbrace{00 \cdots 03}_{m-1 \uparrow} . \\
\text { Let } 2+9(m-1)+5+3=1999 .
\end{array}
$$

So, we only need to take $m=222$.