$10$ distinct integers with sum of any $9$ a perfect square

Assume we have such integers $a_1, \ldots, a_{10}$. Let $a=\sum a_i$. Then we need that the numbers $b_i:=a-a_i$ are squares. In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$\sum_{i=1}^{10} b_i=\sum_{i=1}^{10} (a-a_i)=9a.$$ This is not more than requiring the sum of ten squares to be a multiple of $9$. Note that for $m\in\mathbb Z$ we have $m^2\equiv 0, 1, 4\text{ or }7\pmod 9$. The sum of three numbers $\in\{0,1,4,7\}$ can be any residue class mod $9$. Therefore: Select $7$ arbitrary distinct squares $b_1, \ldots, b_7$. Then select three further distinct squares $b_8,b_9,b_{10}$ such that $b_8+b_9+b_{10}\equiv -(b_1+\ldots+b_7)\pmod 9$. Finally let $a_i=\frac19\sum_{j=1}^{10} b_j - b_i$.

Example: Let $b_i=i^2$ for $i=1, \ldots 7$. Then $-(b_1+\ldots+b_7)\equiv 4\pmod 9$. So we want to obtain $4\pmod 9$ as sum of three numbers $\in\{0,1,4,7\}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$. Then $a=\frac19\sum b_i=54$ and we arrive at $$(a_1, \ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$

In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $\frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case. Here's a strictly very positive example: $$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$