Bumping!

Call the real numbers $a_{-k}<a_{-k+1}< \cdots< a_0< \cdots< a_k$ Call the center of a nice triple the average of the other two in the triplet. Observe $a_i$ can be the center of at most $k-|i|$ triples. (WLOG $i$ is nonnegative, and count the larger elements - there are $k-i$ of them) Notice $$k^2 = \#(\text{nice triples}) \le \sum\limits_{i=-k}^k (k-|i|)=(2k+1)k-\sum\limits_{i=-k}^k |i|=(2k+1)k - 2\sum\limits_{i=1}^k i=(2k+1)k-(k+1)k=k^2$$ Therefore, $a_i$ is exactly the center of at most $k-|i|$ triples. We subtract each element by $a_0$ and divide each element by $a_1$. This way, $a_1=1, a_{-i}=-a_i, a_0=0$. Suppose $2-a_i=a_{x_i}$ for $2\le i\le k$. Note the $x_i$'s are not positive, $x_i<x_j \leftrightarrow i>j$. Furthermore, since $a_i+a_{-i}=0, x_i+i>0$ Since $x_k>-k$, it follows that $\{x_2,\cdots,x_k\}$ is $\{0,-1,\cdots,-(k-1)\}$, minus one element. Call this element $t$ (if $t=-(k-1)$, $a_i$ is just an arithmetic sequence). Then, for all $i<2-t, x_i=2-i$ and for all $i\ge 2-t, x_i=1-i$. This is actually enough to determine the $a_i$'s. We can see for all $i<2-t$, $a_i=i$ by induction on $i$. Since $2-a_{2-t}=a_{t-1}=t-1$, it follows $a_{2-t}=2-(t-1)=3-t$. We can show for all $i>0, a_{1-t+i}=(1-t)+2i$ We can see all elements from $a_k=a_{1-t+(k+t-1)}=(1-t)+2(k+t-1)$ to $-a_k=-(1-t)-2(k+t-1)$ with the same parity as $a_k$ are included in the sequence and no other elements with that parity is included. We can also see all elements from $t$ to $-t$ congruent to $t$ mod 2 are included in the sequence and no other elements with that parity is included. The conclusion follows as those two sequences are disjoint.