The answer is 495.

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Let $k$ be our required number.

Since $k$ is a sum of 9 consecutive positive integers, it can be written as:

$k = x+(x+1)+(x+2)+\ldots+(x+8) = 9x + 36$

Similarly, as $k$ is also a sum of 10 consecutive positive integers, it can be written as:

$k = y + (y+1) + (y+2) + \ldots + (y+9) = 10y + 45$

And as a sum of 11 consecutive positive integers:

$k = z + (z+1) + (z+2) + \ldots + (z+10) = 11z + 55$

Thus we get:

$k=9x+36=9(x+4)$

$k=10y+45=5(2y+9)$

$k=11z+55=11(z+5)$

We see that $k$ has factors $9, 5, 11$ . Thus the smallest number that satisfies the condition to be $k$ must be the smallest number that has all factors $9, 5, 11$ which is $lcm(9, 5, 11)=\boxed{495}$