Language whose intersection with a CFL is always a CFL
Let $L = \{a^n b^n : n \geq 0\}$, and let $L_0$ be an arbitrary context-free language. Define $L_1 = L_0 \cap a^* b^*$, and note that $L_1$ is context-free and $L \cap L_0 = L \cap L_1$. Let $S = \{(i,j) : a^i b^j \in L_1\}$.

According to Parikh's theorem, the set  $S$ is semilinear. The set $D = \{(n,n) \geq 0\}$ is also semilinear (in fact, it is linear). Since the semilinear sets are closed under intersection, $S \cap D$ is also semilinear. Since $S \cap D$ is (essentially) one-dimensional, it is eventually periodic. This shows that there is a finite language $F$, a modulus $m \geq 1$ and a subset $A \subseteq \{0,\ldots,m-1\}$ such that
$$
L \cap L_1 = F \Delta \{ a^n b^n : n \bmod m \in A \},
$$
where $\Delta$ is symmetric difference. It is easy to check that $\{a^nb^n : n \bmod m \in A\}$ is context-free, and so $L \cap L_1$ is context-free.

Summarizing, we have shown that $L$ is a non-regular language which satisfies your condition.