$10$ distinct integers with sum of any $9$ a perfect square Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?

I think the answer is yes. Here is a simple idea:
Consider the system of equations 
$$S-x_i= y_i^2,  1 \leq i \leq 10\,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $\operatorname{rank}(I+A)=1$. This shows that $\lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $\lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $\lambda_1=...=\lambda_{n-1}=-1$ and $\lambda_n=(n-1)$. This shows that $\det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system 
$$S-x_i= y_i^2  1 \leq i \leq 10\,,$$ 
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i \neq j$. Then
$$S-x_i =y_i^2 \,;\, S-x_j=y_j^2 \Rightarrow x_i-x_j=y_j^2-y_i^2 \neq 0 \,.$$
Remark You can easily prove that $\det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
