Note that $n \equiv S(n) \pmod{9}$. This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$. Thus, if $S(n) = 1274$, then $n \equiv 5 \pmod{9}$, and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$. The only answer choice that satisfies $n+1 \equiv 6 \pmod{9}$ is $\boxed{\textbf{(D)} 1239}$.

1. Let $S(n) = 1274$. This means the sum of the digits of $n$ is 1274.
2. When we add 1 to $n$, we need to consider how this affects the sum of the digits. If adding 1 to $n$ does not cause any carries, then $S(n+1) = S(n) + 1 = 1274 + 1 = 1275$.
3. However, if adding 1 to $n$ causes carries, the sum of the digits can change more significantly. For example, if $n$ ends in a sequence of 9's, adding 1 will turn these 9's into 0's and increase the next digit by 1.
4. Consider the case where $n$ ends in $k$ nines. For example, if $n = 999 \ldots 999$ (with $k$ nines), then $n+1 = 1000 \ldots 000$ (with $k$ zeros and a leading 1). The sum of the digits changes from $9k$ to 1.
5. In general, if $n$ has a sequence of $k$ trailing nines, adding 1 will decrease the sum of the digits by $9k$ and increase it by 1 (from the carry). Therefore, $S(n+1) = S(n) - 9k + 1$.
6. We need to find a value of $k$ such that $S(n+1)$ is one of the given options: 1, 3, 12, 1239, or 1265.

Let's check each option:
- For $S(n+1) = 1$:
$$1 = 1274 - 9k + 1 \implies 0 = 1274 - 9k \implies 9k = 1274 \implies k = \frac{1274}{9} \approx 141.56$$
Since $k$ must be an integer, this is not possible.

- For $S(n+1) = 3$:
$$3 = 1274 - 9k + 1 \implies 2 = 1274 - 9k \implies 9k = 1272 \implies k = \frac{1272}{9} \approx 141.33$$
Since $k$ must be an integer, this is not possible.

- For $S(n+1) = 12$:
$$12 = 1274 - 9k + 1 \implies 11 = 1274 - 9k \implies 9k = 1263 \implies k = \frac{1263}{9} \approx 140.33$$
Since $k$ must be an integer, this is not possible.

- For $S(n+1) = 1239$:
$$1239 = 1274 - 9k + 1 \implies 1238 = 1274 - 9k \implies 9k = 36 \implies k = \frac{36}{9} = 4$$
This is possible since $k = 4$ is an integer.

- For $S(n+1) = 1265$:
$$1265 = 1274 - 9k + 1 \implies 1264 = 1274 - 9k \implies 9k = 10 \implies k = \frac{10}{9} \approx 1.11$$
Since $k$ must be an integer, this is not possible.

Therefore, the only possible value for $S(n+1)$ is 1239.

The final answer is $\boxed{1239}$.