Digit sum and alternating sum

Write the number with leading zeros as $6$ digits, and split the digits into those in odd and even positions.

1. Let $A$ be the sum of the digits in positions $1$, $3$, $5$ and $B$ the sum in positions $2$, $4$, $6$; then $A+B=27$, and the number is divisible by $11$ exactly when $A-B$ is a multiple of $11$.
2. Because $A+B$ is odd, $A-B$ is odd, and since $|A-B\mid \le 27$, it must be $11$ or $-11$, giving $(A,B)=(19,8)$ or $(8,19)$.
3. The number of triples of digits from $0$ to $9$ with sum $8$ is $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=45$, and the triples with sum $19$ correspond to them by replacing each digit $d$ by $9-d$, so there are also $45$.
4. Each of the $2$ cases gives $45\times 45=2025$ numbers, so the total is $2\times 2025=4050$.

Answer: $4050$