context free grammar for palindrome: $L_n = \{x \in \Sigma^* | x = ywz, w^R = w, |w| \geq n, |y| = |z| \}$
When $n=1$, the language $L_1$ consists of all words of the form $ywz$, where $|y|=|z|$ and $w$ is a non-empty palindrome. We can generate all non-empty palindromes as follows:
$$
P \to 0P0 \mid 1P1 \mid 0 \mid 1 \mid 00 \mid 11
$$
Given that, we can generate $L_1$ by capturing also the outer part:
$$
S \to 0S0 \mid 0S1 \mid 1S0 \mid 1S1 \mid P
$$
For larger $n$, all we need to change is the "base cases" for $P$.

Further observation reveals that we can actually assume that $|w| \leq 2$ (for $n = 1$), and so we can use alternatively the following productions for $P$:
$$
P \to 0 \mid 1 \mid 00 \mid 11
$$
This allows us to eliminate $P$ from the grammar, reaching the following grammar:
$$
S \to 0S0 \mid 0S1 \mid 1S0 \mid 1S1 \mid 0 \mid 1 \mid 00 \mid 11
$$
These steps can also be extended to larger $n$.