Rewrite the sum as $\sum_{x^{2}+x y+y^{2} \leq 10^{6}}\left(x^{2}+x y+y^{2}\right)$ where the sum is over all pairs $(x, y)$ of integers with $x^{2}+x y+y^{2} \leq 10^{6}$. We can find a crude upper bound for this sum by noting that $x^{2}+x y+y^{2}=\frac{3}{4} x^{2}+\left(\frac{x}{2}+y\right)^{2} \geq \frac{3}{4} x^{2}$ so each term of this sum has $|x| \leq \frac{2}{\sqrt{3}} 10^{3}$. Similarly, $|y| \leq \frac{2}{\sqrt{3}} 10^{3}$. Therefore, the number of terms in the sum is at most $\left(\frac{4}{\sqrt{3}} 10^{3}+1\right)^{2} \approx 10^{6}$. (We are throwing away "small" factors like $\frac{16}{3}$ in the approximation.) Furthermore, each term in the sum is at most $10^{6}$, so the total sum is less than about $10^{12}$. The answer $1 \cdot 10^{12}$ would unfortunately still get a score of 0. For a better answer, we can approximate the sum by an integral: $\sum_{x^{2}+x y+y^{2} \leq 10^{6}}\left(x^{2}+x y+y^{2}\right) \approx \iint_{x^{2}+x y+y^{2} \leq 10^{6}}\left(x^{2}+x y+y^{2}\right) d y d x$. Performing the change of variables $(u, v)=\left(\frac{\sqrt{3}}{2} x, \frac{1}{2} x+y\right)$ and then switching to polar coordinates $(r, \theta)=\left(\sqrt{u^{2}+v^{2}}, \tan ^{-1}(v / u)\right)$ yields $\iint_{x^{2}+x y+y^{2} \leq 10^{6}}\left(x^{2}+x y+y^{2}\right) d y d x =\frac{2}{\sqrt{3}} \iint_{u^{2}+v^{2} \leq 10^{6}}\left(u^{2}+v^{2}\right) d v d u =\frac{2}{\sqrt{3}} \int_{0}^{2 \pi} \int_{0}^{10^{3}} r^{3} d r d \theta =\frac{4 \pi}{\sqrt{3}} \int_{0}^{10^{3}} r^{3} d r =\frac{\pi}{\sqrt{3}} \cdot 10^{12}$. This is approximately $1.8138 \cdot 10^{12}$, which is much closer to the actual answer. (An answer of $1.8 \cdot 10^{12}$ is good enough for full credit.)