Is $\frac{n}{3}$-CLIQUE NP-complete?
Let me explain how to solve this, leaving all details to you.

Given an instance $(G,k)$ of CLIQUE, we want to construct an equivalent instance of $\frac{n}{3}$-CLIQUE, say $G'$. Let us denote by $|G|$ the number of vertices in $G$. There are several cases to consider:


Case 1: $k = |G|/3$. In this case you can take $G' = G$.
Case 2: $k &gt; |G|/3$. In this case you want to increase the number of vertices to $3k$ without changing the size of the maximum clique.
Case 3: $k &lt; |G|/3$. This is the most complicated case. You want to increase the number of vertices to some $n'$ while increasing the size of the maximum clique by some $\Delta$. You are done if you can ensure that $k+\Delta = n'/3$. (In fact, it suffices to have $k+\Delta &gt; n'/3$, since then we can use Case 2.)


You take it from here.