A positive integer is called oddly even if the sum of its digits is even. Find the sum of the first 2013 oddly even numbers.
For convenience, we declare 0 to be oddly even (its digit sum is of course 0), and we sum the first 2014 oddly even nonnegative integers.

Let us look at the oddly even numbers in a given hundred, that is, in the range $100 n$ to $100 n+99$ where $n \geq 0$ is an integer. For each tens digit, there are five possibilities for the units digit, and for each units digit, there are five possibilities for the tens digit. In particular, there are exactly 50 oddly even numbers in this range. When we sum their units digits, we get

$$
5 \cdot 0+5 \cdot 1+\cdots+5 \cdot 9
$$

which is exactly half the sum of the units digits of all integers in the range $[100 n, 100 n+99]$. The same can be said for the tens digits and all preceding digits, so the sum of all oddly even numbers in the range $[100 n, 100 n+99]$ is half the sum of all the integers in this range.

We deduce that the interval $[0,3999]$ has 2000 oddly even numbers and their sum is

$$
\frac{1}{2}(0+1+\cdots+3999)=\frac{1}{2} \cdot \frac{3999 \cdot 4000}{2}=3999 \cdot 1000=3999000 .
$$

We compute the next fourteen oddly even numbers:

4000, 4002, 4004, 4006, 4008, 4011, 4013, 4015, 4017, 4019, 4020, 4022, 4024, 4026.

Their sum can be computed by hand using various expendients, for instance:

$$
\begin{array}{r}
14 \cdot 4000+(0+2+4+6+8+11+13+15+17+19+20+22+24+26) \\
=14 \cdot 4000+[(2+8)+(4+6)+(11+19)+(13+17)+(24+26)+20+15+22] \\
=14 \cdot 4000+[10+10+30+30+20+50+37]=56000+187=56187 .
\end{array}
$$

So the total sum of the first 2013 oddly even numbers is

$$
3999000+56187=4055187
$$