<s>[INST] Finding all non-negative integers solutions to $x_1+x_2+x_3+...+x_6=20$ such that $x_{2n+1} \le x_{2n+2}$ for $0 \le n \le 2$ Solve $x_1+x_2+x_3+x_4+x_5+x_6=20$ such that  $x_{2n+1}\leq x_{2n+2}, 0\leq n \leq2$
Edit : I solved it.
Let $0 \le a,b,c \le 20$ such that $0 \le a+b+c \le 20$
$x_1+x_2+x_3+x_4+x_5+x_6=20 \;and\ x_{2n+1}\leq x_{2n+2}\\
\implies x_2=x_1+a,\;x_4=x_3+b,\;x_6=x_5+c\\
\implies 2(x_1+x_2+x_3)+a+b+c=20\\
\implies 2(x_1+x_2+x_3)=20-a-b-c\;\;\;(1)$ 
So now we need for a fixed values of $a,b,c$ to find the number of non-negative integers solutions to the equation at $(1)$
Now, for all $0 \le i \le 20 $
Let $A_i$ be the set of all non negative solutions for $a+b+c=i$
Let $X_i$ be the set of all non negative solutions for $2(x_1+x_2+x_3)=20-i$
Now,
$2(x_1+x_2+x_3)=20-i \implies (x_1+x_2+x_3)= \frac{20-i}{2}$
And since we are interesting in non-negative integers we can say that 
$20-i\nmid 2 \implies X_i=\emptyset \implies |X_i|=0\;\;\;(2)$ 
Otherwise,
If $i$ is even we will want to compute 
$\left|A_i\right|\cdot \left|X_i\right|$  because for any solution a+b+c=i in $A_i$ there's the corresponding solution in $X_i$
Therefore we have:
$\sum _{i=0}^{20}\:\left(\left|A_i\right|\cdot \left|X_i\right|\right)=\sum _{i=0}^{20}\:\begin{pmatrix}i+3-1\\ \:i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:10-1\end{pmatrix}\\ \text{and from (2)}\\\sum _{i=0}^{20}\:\begin{pmatrix}i+3-1\\ \:i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:10-1\end{pmatrix}= \sum \:_{i=0}^{10}\:\begin{pmatrix}2i+3-1\\ \:\:2i\end{pmatrix}\begin{pmatrix}10-i+3-1\\ \:\:10-i\end{pmatrix}=\\ \sum _{i=0}^{10}\:\left(\frac{\left(2i+2\right)\left(2i+1\right)}{2}\right)\left(\frac{\left(12-i\right)\left(11-i\right)}{2}\right)=9009$
And we've got a palindrome.
 [/INST] I verified your solution a different way:
Let $N(i)$ denote the number of ways to have $x_{1}+x_{2}=i$ with $x_{1}\leq x_{2}$. It is easy to show that $N(i)=\lfloor\frac{i}{2}\rfloor+1$, for $0\leq i$. 
Your problem now reduces to computing the following sum: 
$\sum_{(k_{1},k_{2},k_{3})\in K}N(k_{1})\cdot N(k_{2})\cdot N(k_{3})$
where $K=\{(k_{1},k_{2},k_{3}):k_{1}+k_{2}+k_{3}=20,\ k_{i}\geq 0\}$, i.e. the set of ordered pairs of three non-negative integers summing to 20.
After a bit of sum-manipulating we get the equivalent sum: 
$\sum_{k_{1}=0}^{20}N(k_{1})\sum_{k_{2}=0}^{20-k_{1}}N(k_{2})\cdot N(20-(k_{1}+k_{2}))$
Now throw elegance to the curb, enter the sum into Mathematica, and get 9009 $\checkmark$. 
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