= Solution
Use the density form of the <Szemerédi theorem>: for fixed $\alpha>0$ and length $r$, every sufficiently large subset of $\{1,\ldots,N\}$ of size at least $\alpha N$ contains a nonconstant <arithmetic progression> of length $r$, as established in https://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27132.pdf[Szemerédi's density theorem]. This theorem supplies a progression, not the desired averaging configuration; we now select the actual indices.
Take $r=2k+1$. Write the resulting progression as
$$
a,a+d,\ldots,a+2kd\in A,\qquad d\ge1,
$$
and put $y=a+kd\in A$. If $k=2m$ is even, select $y-d,y+d,\ldots,y-md,y+md$. These are $2m=k$ distinct members of the progression, and each opposite pair sums to $2y$. If $k=2m+1$ is odd, select those $2m$ opposite points together with $y$ itself. All $k$ selected points are distinct, and their sum is again $ky$. The case $k=1$ simply selects $y$.
In both cases the indices stay between $0$ and $2k$ and the difference $d$ is nonzero. Thus the precise answer is
$$
\boxed{x_1,\ldots,x_k\in A\text{ distinct},\qquad\frac{x_1+\cdots+x_k}{k}=y\in A.}
$$
No deletion of diagonal solutions and no multidimensional density theorem is needed. The averaging point is allowed to be one of the selected points, as it is in the odd case.
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