(14) Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=2012+3 n$, find all positive integers $n$ such that the first $n$ terms of the sequence $\left\{a_{n}\right\}$ can be divided into two parts, the sums of which are equal.
(14) $\left\{a_{n}\right\}$ is an arithmetic sequence, $S_{n}=2012 n+\frac{3}{2} n(n+1)$.
To make the sum of the first $n$ terms of the sequence divisible into two equal parts, $S_{n}$ must be divisible by 2.
Therefore, $n \equiv 0(\bmod 4)$ or $n+1 \equiv 0(\bmod 4)$.
(1) If $n \equiv 0(\bmod 4)$, i.e., $n=4 k, k \in \mathbf{N}^{*}$.
Then, from $a_{1}+a_{n}=a_{2}+a_{n-1}=\cdots=a_{2 k}+a_{2 k+1}$, we know:
From these $2 k$ pairs, select $k$ pairs as one part, and the rest as the other part, so that the sums of these two parts are equal.
(2) If $n+1 \equiv 0(\bmod 4)$, i.e., $n=4 k-1, k \in \mathbf{N}^{*}$.
Let the first part have $i$ terms, and the second part have $j$ terms $(i>j, i+j=n)$, with the sums of the two parts being $T_{1}$ and $T_{2}$, respectively.

If $T_{1}=T_{2}$, then $T_{1} \equiv 2012 i \equiv -i(\bmod 3), T_{2} \equiv 2012 j \equiv -j(\bmod 3)$, so $i \equiv j(\bmod 3)$, and $i-j$ is a multiple of 3.

Since $i+j=4 k-1$ is odd, $i-j$ is also odd. Let $i-j=3(2 l+1)$, $l \in \mathbf{N}$, then $i=2 k+3 l+1, j=2 k-3 l-2$.
Because $T_{1} \geqslant a_{1}+a_{2}+\cdots+a_{i}=2012 i+\frac{3}{2}(1+i) i$,
$$
\begin{aligned}
& T_{2} \leqslant a_{i+1}+a_{i+2}+\cdots+a_{4 k-1} \\
= & 2012 j+\frac{3}{2}(i+1+4 k-1) j \\
= & 2012 j+\frac{3}{2}(4 k+i) j,
\end{aligned}
$$

Since $T_{1}=T_{2}$, $n$ must satisfy
$$
2012 i+\frac{3}{2}(1+i) i \leqslant 2012 j+\frac{3}{2}(4 k+i) j,
$$

So $4 k^{2}-8 k \geqslant 2014+9 l^{2}+4033 l+12 l k$,
$4 k^{2}-8 k \geqslant 2014, k \geqslant 24$, i.e., $n=4 k-1 \geqslant 4 \times 24-1=95$.
When $n=95, k=24$,
$$
\sum_{n=50}^{95}(2012+3 n)-\sum_{n=1}^{49}(2012+3 n)=102557-102263=294,
$$
$294 \div 6=49$, so by swapping the 1st term and the 50th term, the sums of the two parts are equal. That is, taking the 2nd to the 50th terms as one part, and the rest as the other part, the sums of the two parts are equal, which is 102410.

When $n>95$ and $n=4 k-1$, $n$ can be expressed as $n=95+(k-24) \times 4$, then the first 95 terms are divided according to $n=95$, and the remaining $(k-24) \times 4$ terms are divided according to (1).

In this way, the sums of the two parts are equal.
In summary, $n \in\left\{n \mid n=4 k, k \in \mathbf{N}^{*}\right\} \cup\{n \mid n=4 k-1, k \geqslant 24$ and $\left.k \in \mathbf{N}^{*}\right\}$.