$\mathcal{G} = \{ v_2 v_4 \ldots v_{k} : v_1 v_2 v_3 v_4 \ldots v_{k-1} v_{k} \in \mathcal{L}, \text{ $k$ even} \} $ is context free language
This answer assumes that $v_i \in \Sigma$ are individual symbols.

You can prove this using closure properties. The advantage is that any class of languages closed under the requisite closure properties will be closed under this operation. Specifically, we will need closure under homomorphism, reverse homomorphism, and intersection with regular language, which are exactly the so-called "full trio".

Let $\Sigma' = \{ \sigma' : \sigma \in \Sigma \}$ be a copy of $\Sigma$. Define homomorphisms $r,d\colon \Sigma \cup \Sigma' \to \Sigma$ by $r(\sigma) = r(\sigma') = \sigma$ and $d(\sigma) = \sigma$, $d(\sigma') = \epsilon$. Then
$$
\mathcal{G} = d(r^{-1}(\mathcal{L}) \cap (\Sigma' \Sigma)^*).
$$

Some families of languages, for example context-sensitive languages, are closed under the so-called "trio", in which homomorphism is replaced with $\epsilon$-free homomorphism (meaning $h(\sigma) \neq \epsilon$ for all letters $\sigma$). We can accommodate these as well with a slightly more complicated argument.

Let $e\colon \Sigma' \times \Sigma \to \Sigma \cup \Sigma$ be given by $e(\sigma',\sigma) = \sigma' \sigma$, and let $p\colon \Sigma' \times \Sigma \to \Sigma$ be given by $p(\sigma',\sigma) = \sigma$. Then
$$
\mathcal{G} = p(e^{-1}(r^{-1}(\mathcal{L}) \cap (\Sigma'\Sigma)^*)).
$$