Integer midpoint

Just to reiterate what I said earlier, if we list the possible coordinate parities of the $3$-tuples, we have

$$\{(e,e,e),(e,e,o),(e,o,e),(e,o,o),(o,e,e),(o,e,o),(o,o,e),(o,o,o)\}$$

So there are $8$ different parity $3$-tuples possible. Also note the addition of even and odd integers;

$$e+e=e,$$ $$o+o=e,$$ $$e+o=o$$

So in order for your midpoint to be an integer, we need to have the points being added to have coordinates of the same parity. Therefore, since we have 9 points, assuming our first 8 points are all of different parity combinations as above, the 9th must be one of these 8 possibilities. If you add up the two points that have the same coordinate parity structure, this will yield an even number, which is divisible by 2. For example, if both ordered triples have coordinate parity structure $(e,e,o)$, then by the midpoint formula,

$$Mid((e,e,o),(e,e,o))=\left(\frac{e+e}{2}, \frac{e+e}{2},\frac{o+o}{2}\right)=\left(\frac{2e}{2}, \frac{2e}{2},\frac{2o}{2}\right)=(e,e,o)$$

And so by the Pigeonhole principle, there is at least 1 midpoint that contains integer coordinates. And as Robert Z noted in his solutions, this can be extended to $n$-tuples as well.

Hint. Read carefully Eleven-Eleven's comment (or take a look at Pigeon-Hole Principle and 2d grid?) and, by using the Pigeonhole Principle, show the following more general statement: in any set of $2^n+1$ points in $\mathbb{Z}^n$ there is at least one pair that has the midpoint in $\mathbb{Z}^n$.