Given $1000000000$ first natural numbers. We change each number with the sum of its digits and repeat this procedure until there will remain $1000000000$ one digit numbers.  Is there more "$1$"-s or "$2$"-s?
1. **Understanding the Problem:**
   We start with the first $1,000,000,000$ natural numbers and repeatedly replace each number with the sum of its digits until all numbers are reduced to single-digit numbers. We need to determine whether there are more $1$'s or $2$'s among these single-digit numbers.

2. **Sum of Digits and Digital Root:**
   The process of repeatedly summing the digits of a number until a single-digit number is obtained is equivalent to finding the digital root of the number. The digital root of a number $n$ can be found using the formula:
   \[
   \text{Digital Root}(n) = 1 + (n - 1) \mod 9
   \]
   This formula ensures that the digital root is a number between $1$ and $9$.

3. **Distribution of Digital Roots:**
   For the first $1,000,000,000$ natural numbers, we need to determine the frequency of each possible digital root (from $1$ to $9$). Since the digital root function cycles every $9$ numbers, each of the digits $1$ through $9$ will appear with equal frequency in any complete cycle of $9$ numbers.

4. **Counting the Frequencies:**
   Since $1,000,000,000$ is a multiple of $9$, we can divide it into $1,000,000,000 / 9 = 111,111,111$ complete cycles of $9$ numbers. In each cycle, each digit from $1$ to $9$ appears exactly once.

5. **Conclusion:**
   Therefore, in $1,000,000,000$ numbers, each of the digits $1$ through $9$ will appear exactly $111,111,111$ times. This means there are exactly $111,111,111$ occurrences of the digit $1$ and exactly $111,111,111$ occurrences of the digit $2$.

\[
\boxed{\text{There are equal numbers of } 1 \text{ and } 2.}
\]