Because there are $15$ ones, therefore, regardless how to insert the "$+$" operators, the sum will always be a multiple of $3$. In order to make the sum a multiple of $30$, it is sufficient to make the sum a multiple of $10$, i.e. the unit digit of the sum must be $0$. Because $15$ is larger then $10$ but less than $20$, so we must have exactly $10$ summands or $9$ addition operators. Then the problem becomes how to place $9$ additional operators in $14$ available spaces. Hence, the answer is $$\binom{14}{9}=\boxed{2002}$$