Recall that a number is divisible by $9$ if and only if the sum of its digits is also divisible by $9.$

This means that looking at $S(n)$ mod $9$ would also give us $n$ mod $9.$

Let us prove this. If we add $x$ to $n$ without carrying, it is clear that the sum of the digits increases by $x$ and that $n$ itself increases by $x.$

This would increase both their values mod $9$ by $x.$

Now, if it does carry, we would be subtracting $10$ from some digit and adding on $1$ to the next digit.

This would keep the value mod $9$ constant. We did, however, add $x$ in there, so the value mod $9$ still increased by $x.$

These are the only two cases, and in both we have shown that the value mod $9$ for both $n$ and $S(n)$ increased by $x.$

Therefore, we have that $$n \equiv S(n) \equiv 5 \pmod 9.$$

From this, we can see that $$\begin{aligned} n + 1 &\equiv 6 \\ &\equiv S(n + 1) \pmod 9. \end{aligned}$$

The only answer choice that leaves a remainder of $6$ when divided by $9$ is $1239.$

This value is attainable. Take $n$ to consist of $137$ digits $9,$ followed by a digit $5,$ followed by four more digits $9.$ Then $$\begin{aligned}S(n)&=137\cdot9+5+4\cdot9\\&=1274.\end{aligned}$$ Adding $1$ changes the final four $9$s to $0$s and changes the $5$ to $6,$ so $S(n+1)=137\cdot9+6=1239.$

Thus, D is the correct answer.