jmc

Note that $$2010=a^2-b^2+c^2-d^2=(a-b)(a+b)+(c-d)(c+d).$$If either $a-b>1$ or $c-d>1,$ then $$(a-b)(a+b)+(c-d)(c+d)>(a+b)+(c+d)=2010,$$which is a contradiction. Therefore, we must have $a-b=1$ and $c-d=1.$ In other words, setting $b=a-1$ and $d=c-1,$ we have $$a+b+c+d=2a+2c-2=2010⟹a+c=1006,$$and we must have $a\ge c+2,$ $c\ge 2.$ The pairs $(a,c)$ satisfying these conditions are $(a,c)=(1004,2),(1003,3),\dots ,(504,502),$ which makes $\boxed{501}$ possible values for $a.$