The first sequence has the form $13+4 m$, while the second has the form $11+5 n$. These are equal when $5 n-4 m = 2$, or, for any nonnegative integer $k$, $m=2+5 k$ and $n=2+4 k$ for members of both sequences. Thus, the form of the sequence is $a_k = 11+5 (2+4 k) = 21+20 k$. The sum of the first 100 of these terms is
$$\sum_{k=0}^{99} (21+20 k) = 100 \cdot 21 + \frac12 (20) (100)(99) = 101100$$
ADDENDUM
Looks like you were confused about how to express the common term $a_k$. Here, I just made a table of ordered pairs $(m,n)$ that satisfied $5 n-4 m=2$, which isn't hard. Once you see the pattern, the above expressions jump out.
Well you can't reach $199$, but you can reach $198$. I'll prove both:
In order to reach $199$, we need to reach $0$ with the first $98$ numbers.
But $\dfrac {98(98+1)}2=49\times 99$ is odd, so you can't split the list $\{1,\dots,98\}$ into 2 parts with equal sum.
[EDIT2: An easier way to see this is to notice the parity of the odd sum and $1+2+\dots+100$ must be the same; changing signs do not change the parity.]
Here is a way to reach $198$:
$(1,96,2,95,\dots,48,49,97,99,98,100)$
The first $96$ numbers cancel out. $97+99-98+100=198$.
EDIT: To see the relation to A047415, we consider this:
If the sum of the first $(n-2)$ numbers is even, we can split the first $(n-2)$ numbers into two parts of equal sum. Manipulating the order of our sum will give $0$, so we can reach $(n - 1) + 1 = 2n-1$.
If the sum of the first $(n-2)$ numbers is odd, the sum of the first $(n-4)$ numbers is even. We can split the first $(n-4)$ numbers into two parts of equal sum. Manipulating the order of our sum will give $0$, so we can reach $(n-3)+(n-1)-(n-2)+n=2n-2$.
For $n = 4k+r$, the sum of the first $(n-2)$ numbers is $\dfrac {(n-2)(n-1)}2 = \dfrac {(4k+r-2)(4k+r-1)}2$, which is even for $r=1,2$, odd for $r=0,3$.
Hence:
$$4k\mapsto 8k-2$$$$4k+1\mapsto 8k+1$$$$4k+2\mapsto 8k+3$$$$4k+3\mapsto 8k+4$$
which is $1,3,4,6 \pmod 8$, in that order.
Best Answer
If you are familiar with unions and intersections of sets , then it is not a difficult problem.
Your answer should be: Total sum-(sum of multiple of $3 +$ sum of multiple of $7 -$ sum of multiple of $21)$
Since $21$ is LCM of $3$ & $7$