14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=2012 ?
$$
14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.

Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$

Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is an integer greater than 1.
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Clearly, $(n, 3 n+1)=1$.
Hence, $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Also, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$, which is a contradiction.
Therefore, $f(n) \neq 2$.
When $n=8$,
$$
3 n^{2}+n+1=201 \text {. }
$$

Thus, $f(8)=3$.
In summary, the minimum value of $f(n)$ is 3.
(2) When $n=2 \times 10^{k}-1$, we have
$$
\begin{array}{l}
3 n^{2}+n+1=12 \times 10^{2 k}-10 \times 10^{k}+3 \\
=11 \underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 03}_{k \uparrow} .
\end{array}
$$

Thus, $f(n)=1+1+9(k-1)+3$
$$
=9 k-4\left(k \in \mathbf{N}_{+}\right) \text {. }
$$
(3) There exists a positive integer $n$ such that
$$
f(n)=2012 \text {. }
$$

In fact, let $n=2 \times 10^{k}-1$. Then $f(n)=9 k-4$.
Let $9 k-4=2012$. Solving for $k$ gives $k=224$.
Therefore, taking $n=2 \times 10^{224}-1$, we get
$$
f(n)=2012 \text {. }
$$
(Zhang Shengchun provided)