Write the two arithmetic progressions explicitly

First AP: $3,7,11,15,\ldots,403$ This has first term $a_1=3$ and common difference $d_1=4$. So its general term is $3+4k \quad (k\ge 0).$

Second AP: $2,5,8,11,\ldots,404$ This has first term $a_2=2$ and common difference $d_2=3$. So its general term is $2+3m \quad (m\ge 0).$

Find the common terms

A common term must satisfy $x\equiv 3 \pmod{4}$ and $x\equiv 2 \pmod{3}.$

Let us check numbers in the first AP: $3,7,11,15,19,\ldots$ Among these, those that are $\equiv 2 \pmod{3}$ are: $11,23,35,47,\ldots$

So the common terms themselves form an AP.

Find the common difference of the common-term AP

Since one AP has difference $4$ and the other has difference $3$, the common terms repeat every $\operatorname{lcm}(4,3)=12.$

Hence the common terms are: $11,23,35,47,\ldots$ with common difference $12$.

Find the last common term

The common terms must lie within both given APs. The first AP ends at $403$, the second at $404$. So the largest possible common term is at most $403$.

Now check the sequence $11+12n \le 403.$ So $12n \le 392$ $n \le \frac{392}{12}=32\frac{2}{3}.$

Thus the largest integer $n$ is $32$.

Therefore the last common term is $11+12\cdot 32=11+384=395.$

Find the number of common terms

The common AP is $11,23,35,\ldots,395.$

Number of terms: $n=\frac{395-11}{12}+1=\frac{384}{12}+1=32+1=33.$

Find the sum of these common terms

Sum of an AP is $S=\frac{n}{2}(\text{first term}+\text{last term}).$

Therefore, $S=\frac{33}{2}(11+395)$ $=\frac{33}{2}\cdot 406$ $=33\cdot 203$ $=6699.$

Comparison with stored answer

Our derived answer is $\boxed{6699}.$

This matches the stored correct answer.