For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$
When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$

When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$

Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$

If there exists an integer $n$ such that $f(n)=2014$, then by $2014 \equiv 1(\bmod 3) \Rightarrow 3 \mid n$. Hence $n^{2}+3 n+1 \equiv 1 \equiv f(n)(\bmod 9)$. But $f(n) \equiv 7(\bmod 9)$, which is a contradiction.
Therefore, there does not exist an integer $n$ such that $f(n)=2014$.
When $k>2, n=10^{k}-3$,
$$
\begin{array}{l}
n^{2}+3 n+1=\left(10^{k}-3\right)^{2}+3\left(10^{k}-3\right)+1 \\
=\underset{k-1 \uparrow}{99 \cdots 9} 7 \underset{k-1 \uparrow}{00 \cdots 01} .
\end{array}
$$

Since $2015=9 \times 223+8$, when $k=224$, i.e., $n=10^{224}-3$, $f(n)=2015$.
(Song Hongjun, Zhejiang Fuyang High School, 311400)