Let the numbers be $a_1 < a_2 < \dots < a_{2k+1}$ in increasing order. First notice that the numbers $a_1, a_2, \dots, a_k, a_{k+1}, a_{k+2}, \dots, a_{2k+1}$ can be the middle element of at most $0, 1, 2, \dots, k-1, k, k-1, \dots, 0$ nice triples, respectively. Therefore, in total we have at most $0 + 1 + 2 + \dots + k - 1 + k + k - 1 + \dots + 0 = k^2$ nice triples, and the equality holds when we reach all above upper bounds. In particular, $a_{k+1}$ has to be the middle element of $k$ nice triples, meaning that the set is symmetric with respect to $a_{k+1}$.

We use induction to prove a stronger statement, namely we can divide the numbers into two arithmetic progressions with equal ratios, which are both symmetric with respect to $a_{k+1}$. For $k = 1$ two subsets $\{a_1, a_3\}$ and $\{a_2\}$ work.

Now for the numbers $a_1 < a_2 < \dots < a_{2k+1}$ with $k^2$ nice triples, we know that $a_1, a_{k+1}, a_{2k+1}$ form a nice triple. So $a_1$ appears in at most $k$ nice triples (the middle element for such triples is not bigger than $a_{k+1}$), also $a_{2k+1}$ appears in at most $k$ nice triples (the middle element for such triples is not smaller than $a_{k+1}$), with one nice triple in common. Therefore, in total there are at most $k + k - 1 = 2k - 1$ nice triples having $a_1$ or $a_{2k+1}$ (or both). Therefore, there are at least $k^2 - (2k - 1) = (k - 1)^2$ nice triples all in the set $\{a_2, a_3, \dots, a_{2k}\}$. Thus the equality holds for all above bounds, and the induction hypothesis shows that one can split the set $\{a_2, a_3, \dots, a_{2k}\}$ into two arithmetic progressions with equal ratios, say $d$, which are both symmetric with respect to $a_{k+1}$.

Now consider the numbers $a_k, a_{k+1}, a_{k+2}$ forming a nice triple while at least two of them are consecutive elements of an arithmetic progression with ratio $d$. This means that the difference between $a_k, a_{k+1}$ is $d/2$ or $d$, the latter case is forcibly the case that the numbers $a_2, a_3, \dots, a_{2k}$ form an arithmetic progression with ratio $d$.

On the other hand, $a_1$ forms a nice triple with $a_{k+1}, a_{2k+1}$, and also with $a_k$ and another number, say $a_t$. So the difference $a_{2k+1} - a_t$ is twice the difference $a_{k+1} - a_k$, which is $d/2$ or $d$ as claimed above. In the first case, $a_{2k+1} - a_t = d$ which states that $a_{2k+1}$ is the next element in the arithmetic progression having $a_t$, (also $a_1$ is an element of such progression because both arithmetic progressions are symmetric with respect to $a_{k+1}$). In the second case, $a_2, a_3, \dots, a_{2k}$ is an arithmetic progression with ratio $d$ and we have $a_{2k+1} - a_t = 2d$, so $t = 2k$. In this case the two subsets $\{a_1, a_2, a_4, \dots, a_{2k-2}, a_{2k}, a_{2k+1}\}$ and $\{a_3, a_5, \dots, a_{2k-1}\}$ are arithmetic progressions with ratio $2d$, both symmetric with respect to $a_{2k+1}$. This concludes the induction statement in both cases. ■