Too many unknowns!

Observe that the arithmetic mean of all these four groups of terms $\large(a_1,a_7,a_{11},a_{17}),(a_2,a_6,a_{12},a_{16}),(a_3,a_5,a_{13},a_{15}),(a_4,a_8,a_{10},a_{14})$ is the same and it is $\large{(a_9)}$ . Thus we have:

$\large{\begin{aligned} (a_1+a_7+a_{11}+a_{17}) &=(a_2+a_6+a_{12}+a_{16}) \\ =(a_3+a_5+a_{13}+a_{15}) &=(a_4+a_8+a_{10}+a_{14})=4\times a_9= 4\times 257 \end{aligned}}$

Now add all the equations to get the following:

$\large{\begin{aligned} x_1(a_1+a_7+a_{11}+a_{17}) &+x_2(a_2+a_6+a_{12}+a_{16}) \\ +x_3(a_3+a_5+a_{13}+a_{15}) &+ x_4(a_4+a_8+a_{10}+a_{14}) =11506\\ \Rightarrow x_1(4\times 257)+x_2(4\times 257) &+x_3(4\times 257)+x_4(4\times 257) = 11506\end{aligned} \\ \Rightarrow (4\times 257)(x_1+x_2+x_3+x_4) =11506 \\ \Rightarrow x_1+x_2+x_3+x_4 = \dfrac{11506}{4\times 257} \approx 11.192 \\ \Rightarrow \lfloor x_1+x_2+x_3+x_4 \rfloor = \Large{\boxed{11}}}$

Make a matrix AX=B, apply the row operation, R 1=R 1+R 2+R 3+R 4 , which will give you in first row all four elements as: 4a 1 + 32d where d is the common difference of successive members and which can also be written as 4 (a 1 +8d) or 4a 9. Apply now R 1=R 1/4a 9 . Hence x 1 + x 2 + x 3 + x 4 =11506/4a 9 = 11.2 . hence the answer is 11

11.192607003891 regardless of changes for x1 + x2 + x3 + x4.

As $a_1, a_2, a_3, \ldots, a_9, \ldots, a_{15}, a_{16}, a_{17}$ form an arithmetic progression.

So, any $a_{n}$ could be written in terms of $a_1$ as $a_{n} = a_1 + (n-1)r$ , for $n \in \mathbb{N}$ and $n > 0$ .

In fact, as an arithmetic progression, it could be written in terms of any term, as $a_{n} = a_{n-1} + r$ , for $n \in \mathbb{N}$ and $n > 0$ . So $a_{10} = a_9 + r$ , and $a_9 = a_8 + r \Rightarrow a_8 = a_9 - r$ .

Generalizing all terms in terms of $a_9$ , the arithmetic progression would be written as $a_{n} = a_9 + (n - 9)r$ , for $n \in \mathbb{N}$ and $n > 0$ .

As $a_9 = 257$ , so $a_{n} = 257 + (n - 9)r$ , for $n \in \mathbb{N}$ and $n > 0$ .

Now, let's do equation 2 + equation 3, substituting terms for their $a_9 = 257$ based definitions (could be equation 1 + equation 4 too, anyway):

Equation 2: $(257 - 2r)x_1 + (257 - 3r)x_2 + (257 - 4r)x_3 + (257 - r)x_4 = 2341$

Equation 3: $(257 + 2r)x_1 + (257 + 3r)x_2 + (257 + 4r)x_3 + (257 + r)x_4 = 3412$

Equation 2 + Equation 3: $(257- 2r+257+2r)x_1 + (257-3r+257+3r)x_2 + (257-4r+257+4r)x_3 + (257-r+257+r)x_4 = 2341 + 3412$

$(2 \cdot 257) (x_1 + x_2 + x_3 + x_4) = 5753$

$(x_1 + x_2 + x_3 + x_4) = \frac {5753}{514}$

$(x_1 + x_2 + x_3 + x_4) = 11.19260\ldots$

$\lfloor x_1 + x_2 + x_3 + x_4 \rfloor = \boxed{11}$