The sum of the first $1$ numbers is $1$
The sum of the next $2$ numbers is $2 + 1$
The sum of the next $3$ numbers is $2 + 2 + 1$
In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers $1234$ is between.  By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$, we find:
$f(50) = 1275$
$f(49) = 1225$
Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number.  Then, we can manually tack on the remaining $2$s to hit $1234$.
We want to find:
$\sum_{n=1}^{49} 1 + 2(n-1)$
$\sum_{n=1}^{49} 2n - 1$
$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$
$2 \sum_{n=1}^{49} n - 49$
$2\cdot \frac{49\cdot 50}{2} - 49$
$49^2$
$2401$
Thus, the sum of the first $1225$ terms is $2401$.  We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{B}$.
Note:  If you notice that the above sums form $1 + 3 + 5 + 7... + (2n-1) = n^2$, the fact that $49^2$ appears at the end should come as no surprise.
The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$