Think!

$x_1 , x_2 , x_3 , ......... , x_{10}$ these integers are in $3k+1$ or $3k+2$ form.

$(3k+1)^2=3m+1$

$(3k+2)^2=3n+1$

Thus the all ten integers are in $3k+1$ form

Their sum leaves remainder of $10*1$ getting divided by 3.

The answer is $\boxed{1}$

Given that none of $x_{1},x_{2},x_{3},\ldots,x_{10}$ are divisible by 3, we know that for each $i = 1,2,3,\ldots,10$ , $x_{i}\not\equiv 0 \;\left(mod \; 3\right)$ . Then either $x_{i} \equiv 1 \;\left( mod \; 3 \right)$ or $x_{i} \equiv 2 \;\left(mod \; 3 \right)$ . If $x_{i} \equiv 1 \;\left(mod \; 3\right)$ , then $x_{i}^{2}\equiv1\;\left(mod\;3\right)$ . If instead $x_{i}\equiv2\;\left(mod\;3\right)$ , then $x_{i}^{2}\equiv4\equiv1\;\left(mod\;3\right)$ . In either case, $x_{i}^{2}\equiv1\;\left(mod\;3\right)$ , so $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots+x_{10}^{2}\equiv10\equiv1\;\left(mod\;3\right)$ . Thus the remainder is 1.