Counting is hard!

No matter how you put the +'s, the number would be divisible by $3$ . Just need to check the divisibility by $10$ . for that, you need exactly ten numbers (or a multiple of ten) that ends in $1$ . So, ten of the fifteen is fixed and the remaining five $1$ 's should be placed in front of the fixed one. This would be a "stars and bars" problem $x_1+\dots+x_{10}=5$ , with the solution $\binom{5+10-1}{5}=\binom{14}{5}=2002$

For the sum of the 1's-integers to be divisible by 30. It must be divisible by 3 and 10. Since the digital sum of the sum of 15 digits of 1 is 15 which is divisible by 3 and hence the sum is divisible by 3. For an 15-digit 1's-integer to be divisible by 10, it must be divided into 10 smaller 1's-integers. To divide 15 digits of 1 into 10 parts is similar to putting 9 +'s in 14 gaps between 1's. The number of ways to do that is given by $N = \binom {14}9 = \boxed{2002}$