How many 8-digit numbers are there whose decimal notation is of the form $a b 2019 c d$ with $a>0$, and which are divisible by 360?
To start, a solution number must be divisible by 10, so we must have $d=0$, and the number written as $a b 2019 c$ must be divisible by 36, hence by 4. The number written as $a b 20100$ is divisible by 100, thus by 4, so the number written as $9 c$ must also be divisible by 4, so $c$ is 2 or 6.

Finally, the number must be divisible by 36, hence by 9, so the sum of its digits must be divisible by 9, so $a+b+2+0+1+9+c$ is divisible by 9, so $a+b+c+3$ is divisible by 9, so $a+b+c$ is 6 or 15 or 24 (the sum of three digits cannot exceed 27).

If $c=2$, then $a+b+c \leqslant 2+9+9=20$, so $a+b+c$ is 6 or 15, so $a+b$ is 4 or 13, which gives the following solutions:

$$
49201920,58201920,67201920,76201920,85201920,94201920.
$$

If $c=6$, since $a>0$, we have $a+b+c>6$ so $a+b+c$ is 15 or 24, so $a+b$ is 9 or 18, which gives the following solutions:

18201960, 27201960, 36201960, 45201960, 54201960, 63201960, 72201960, 81201960, 90201960,

99201960.

Therefore, we have a total of 20 solutions.