Answer: 495

Small Hint:

A sum of an odd number of consecutive integers is divisible by the number of terms

Big Hint:

A sum of ten consecutive integers is congruent to $5\pmod {10}$

Solution:

The sums of $9$ and $11$ consecutive integers are divisible by $9$ and $11,$ so the desired number is a multiple of $99.$ A sum of $10$ consecutive integers has the form $10a+45,$ hence is congruent to $5\pmod {10}.$ The first multiple of $99$ ending in $5$ is $5\cdot99=495,$ and each of the three required representations then exists.