How to derive an upper bound on the minimum number $n(k,d)$ of lattice points in $d$dimensions such that there are some $k$ of these points which have a lattice point centroid.

4$\begingroup$ Finding $n(2,3) = 9$ was a Putnam problem in the early 1970's. $\endgroup$ – Mark Fischler Aug 24 '17 at 18:14

$\begingroup$ $n(k,d)=k$ as written. Clarification of the question needed. $\endgroup$ – js21 Aug 25 '17 at 14:04
The trivial inequality $$n(k, d) \leq k \times n(k, d  1)  (k  1)$$ gives a trivial upper bound: $$n(k, d) \leq (k 1) k^d + 1.$$
An easy lower bound is: $$n(k, d) \geq (k  1) 2^d + 1.$$ This is proved by choosing the following lattice points: for every vector in $\{0, 1\}^d \subseteq (\mathbb{Z}/k\mathbb{Z})^d$, repeat the vector $k  1$ times. One then proves easily by induction on $d$ that there are no $k$ vectors among them which have a lattice point centroid.
The above two inequalities determine the value of $n(2, d)$ for every $d$, namely $n(2, d) = 2^d + 1$.
Interestingly, in the case $d = 1$, one has $n(k, 1) = 2k  1$ (ErdosGinzburgZiv theorem, c.f. GTM 165, section 2.4), which suggests that the lower bound $(k  1) 2^d + 1$ might be a good guess for the exact value.
Also, the case where $k = p$ is prime and $d = 2$ is already a conjecture (according to encyclopedia of mathematics): it is conjectured that $n(p, 2) = 4 p  3$ in this case, coinciding with the lower bound.

2$\begingroup$ Case $d = 1$ is exactly ErdösGinzburgZiv theorem. $\endgroup$ – Aleksei Kulikov Aug 26 '17 at 15:28

$\begingroup$ @AlekseiKulikov Thanks! I updated the answer. $\endgroup$ – WhatsUp Aug 27 '17 at 2:20