Is it possible to create a CFG with succesive palindromes? In one of my homework I am requested to find a Context-free-grammar (CFG) and a push down automaton (PDA) for the following language:

$L = \{x_1\#x_2\#...\#x_k | k \geq 2, \text{ each } x_i \in \{a, b\}^*, \text{ and for some } i \text{ and } j, x_i=x_j^\mathcal{R}\}$

My problem is that the statement 

$\text{ ... and for some } i \text{ and } j, x_i=x_j^\mathcal{R}\}$, 

i.e. any two pairs in the sequence must be each others reverses, forces us to make all $x_i$'s palindromes, as that is the only way any two are guaranteed reverses of each other.

If that interpretation is correct, I think the problem is impossible to solve with a Context free grammar, or not? This leads me to believe I interpreted the statement wrong and it means something else, but I can't figure out what.
A reference to $j$ is missing. My guess is that it should read $x_i=x_j^{\mathcal R}$. This means that some pair $x_i$, $x_j$ should be each others reverse, or one string should be a palindrome. This can be done with a CFG.

(after correction)
It does not say that for each $i$ there should be a $j$, it just states that some $i,j$ exist. One pair will do.  If the word "any" would have been used then this would probably change the meaning, although "any" sometimes is ambiguous.