(14) 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 $n=2 \cdot 10^{k}-1$ (where $k$ is a positive integer), 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, hence $f(n) \neq 1$.

Assume $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digit being 1, and the rest of the digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$. Clearly, when $k=1$, $n$ does not exist, so $k$ is an integer greater than 1.
Thus, $n(3 n+1)=2^{k} \cdot 5^{k}$.
It can be proven that $n$ and $3 n+1$ are coprime. (Because: Suppose $n$ and $3 n+1$ have a common divisor $m$ greater than 1, let $n=p m, 3 n+1=q m, (p, q, m \in \mathbf{N}^{*})$, then $3 p m + 1 = q m$, i.e., $(q-3 p) m = 1$. Therefore, $m=1, q-3 p=1$. This contradicts $m>1$. Hence, $n$ and $3 n+1$ are coprime.)

So $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k},\end{array}\right.$, then $3 n+1 \leqslant 4 n = 4 \cdot 2^{k} < 5^{k}$, which is a contradiction. Therefore, $f(n) \neq 2$.
When $n=8$, $3 n^{2}+n+1=201$, so $f(8)=3$.
In summary, the minimum value of $f(n)$ is 3.
(2) If $n=2 \times 10^{k}-1$, then
$$
3 n^{2}+n+1=12 \times 10^{2 k}-10 \times 10^{k}+3=11 \underbrace{99 \cdots 9900 \cdots 00}_{k-1} 3 .
$$

So $f(n)=1+1+9(k-1)+3=9 k-4$ (where $k$ is a positive integer).
(3) There exists a positive integer $n$ such that $f(n)=2012$.
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 $n=2 \times 10^{224}-1$ yields $f(n)=2012$.