Multiples of 3.

Since all of the numbers have a common factor of 3, we can re-write this sum as $3(1+2+3+...+31+32+33)=x$ Now, we can just find the 33rd triangle number, which can be found using the formula: $3(\frac{33(33+1)}{2})$ This can be simplified to the expression of $99\times17$ which is equivalent to $\boxed{1,683}$ .

Relevant wiki: Arithmetic Progressions

It is an arithmetic progression with a common difference of $3$ . So the number of terms is $\dfrac{99}{3}=33$ . And the desired sum is

$x=\dfrac{33}{2}[2(3)+(33-1)(3)]=\boxed{1683}$