Sum up those natural numbers less than 1980, in which the sum of the digits is even!
I. solution. If we add a thousand to the natural number $n$ which is less than a thousand, the sum of its digits increases by one. Taking 2000 instead of 1980, the sum thus includes exactly one of the numbers $n$ and $1000+n$. Therefore, the

$$
A=0+1+2+\ldots+999=500 \cdot 999
$$

sum is exactly $1000 B$ less than the sum of those natural numbers less than 2000, in which the sum of the digits is even, where $B$ is the number of natural numbers less than a thousand, in which the sum of the digits is odd.

If we subtract the natural number $n$ less than a thousand from 999, the subtraction can be done digit by digit, with no remainder anywhere. The sum of the corresponding digits is therefore 9 for all three place values (if we consider numbers less than 100 as three-digit numbers by adding 0 digits). Therefore, exactly one of the numbers $n$ and 999 - $n$ has an odd sum of digits, so $B=500$.

Finally, let's add up those natural numbers $n$ for which $1980 \leq n < 2000$, and the sum of the digits is even:

$$
\begin{aligned}
C=1980+1982+1984+1986+ & 1988+1991+1993+1995+1997+1999= \\
& =19895
\end{aligned}
$$

The sought sum is therefore

$$
S=A+1000 B-C=500 \cdot 999+500 \cdot 1000-19895=979605
$$

István Molnár (Nyíregyháza, Krúdy Gy. Gymnasium, 3rd grade)

II. solution. Divide any number $n$ by 20, and let the quotient be $h$ and the remainder $m$:

$$
n=20 h+m, \quad 0 \leq m<20
$$

If in this relationship we write the sum of the digits of each of the three numbers, the relationship remains valid. The last digit of the number $n$ is indeed equal to the last digit of $m$, and the last digit of the number $H=20 h$ is 0. The second last digit of $n$ is divided between the second last digits of $H$ and $m$: if this digit is odd, say $2 k+1$, then the second last digit of $H$ is $2 k$, and that of $m$ is 1, otherwise the second last digits of $n$ and $H$ are equal, and that of $m$ is 0.

If the remainder $m$ runs from 0 to 20, and $H$ is constant, in half of the cases, the

$$
0,2,4,6,8,1113,15,17,19
$$

numbers will have an even sum of digits. The sum of these is equal to the sum of the other possible remainders. Therefore, the sum of those numbers $n$ for which $H=20 h \leq n<H+20$, and the sum of the digits of $n$ is even, is equal to half the sum of the numbers between $H$ and $H+19$. Since 1980 is divisible by 20, this implies that the sought sum is equal to half the sum of the natural numbers less than 1980:

$$
S=0+1+2+\ldots+1979=\frac{990 \cdot 1979}{2}=979605
$$