Mathematical Olympiad Rioplatense

Denote the digit sum of $a$ by $S(a)$. Add a card with $0$ and assume $S(0)=0$. Among $0,1,\dots ,999$ there are $500$ numbers $a$ with $S(a)$ odd and $500$ with $S(a)$ even. Indeed $0,1,\dots ,999$ can be divided into $500$ pairs $(a,b)$ with sum $999$ of every pair. There is no carryover in the addition $a+b=999$, so $S(a)+S(b)=S(999)=27$ which is an odd number. Hence $S(a)$ and $S(b)$ have different parity for every pair $(a,b)$, as needed. Let $X$ ($Y$) be the sum of the $500$ numbers with odd (even) digit sum among $0,1,\dots ,999$.

For $a\in \{0,1,\dots ,999\}$ we have $S(1000+a)=S(a)+1$, hence $S(1000+a)$ and $S(a)$ have different parity. So $\{1000,1001,\dots ,1999\}$ contains $500$ numbers $b$ with $S(b)$ odd. They are obtained from the numbers $a\in \{0,1,\dots ,999\}$ with $S(a)$ even by adding $1000$. It follows that the numbers with odd digit sum in $\{0,1,\dots ,1999\}$ have sum $X+Y+500\cdot 1000$. Since $X+Y=0+1+\cdots +999=500\cdot 999$, the numbers in $[1,1999]$ contribute $500\cdot 1999$ to the sum we are looking for. There remain $2000,2001,\dots ,2010$. Of them the ones with odd digit sum are $2001,2003,2005,2007,2009,2010$; they add up to $12035$. So the final answer is $500\cdot 1999+12035=1011535$.