THE DANUBE MATHEMATICAL COMPETITION

The answer is in the affirmative. The terms of the sum can be: $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1$, $(\sqrt{3}+\sqrt{2}{)}^2=5+2\sqrt{6}$ or $(\sqrt{3}-\sqrt{2}{)}^2=5-2\sqrt{6}$. If there are $a$ terms equal to $1$, $b$ terms equal to $5+2\sqrt{6}$ and $c$ terms equal to $5-2\sqrt{6}$, then $a,b,c$ need to satisfy $a+b+c=1008$, $a+(5+2\sqrt{6})b+(5-2\sqrt{6})c=2016$. The last equality can be written $a+5b+5c-2016=\sqrt{6}(2c-2b)$. As $\sqrt{6}$ is irrational, it follows that $b=c$ and $a+5b+5c=2016$. Finally we obtain $a=756,b=c=126$.