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 $X+Y+500\cdot1000$. Since $X+Y = 0+1+\dots+999 = 500\cdot999$, 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 $500 \cdot 1999 + 12035 = 1011535$.