All-Soviet-Union Mathematical Olympiad

Solution:
$1980 = 2^{2} 3^{2} 5 \cdot 11$. $N$ is obviously divisible by $2^{2}$ and $5$. The digits in odd position are $9 + (0 + 1 + 2 + \ldots + 9) + (0 + 1 + 2 + \ldots + 9) + \ldots + (0 + 1 + 2 + \ldots + 9) + 0 = 9 + 6 \cdot 45 = 279$. The digits in even position are $1 + (2 + 2 + \ldots + 2) + (3 + 3 + \ldots + 3) + \ldots + (7 + \ldots + 7) + 8 = 9 + 10(2 + 3 + \ldots + 7) = 279$. So the sum of the digits of $N$ is $2 \cdot 279$ which is divisible by $9$. Hence $N$ is divisible by $9$. The difference between the odd and even sums is $0$, which is divisible by $11$, so $N$ is divisible by $11$.