Consider the number obtained by writing the numbers $1,2,\ldots,1990$ one after another. In this number every digit on an even position is omitted; in the so obtained number, every digit on an odd position is omitted; then in the new number every digit on an even position is omitted, and so on. What will be the last remaining digit?
1. **Calculate the total number of digits in the sequence from 1 to 1990:**
   - Numbers from 1 to 9: \(9\) numbers, each with 1 digit.
   - Numbers from 10 to 99: \(99 - 10 + 1 = 90\) numbers, each with 2 digits.
   - Numbers from 100 to 999: \(999 - 100 + 1 = 900\) numbers, each with 3 digits.
   - Numbers from 1000 to 1990: \(1990 - 1000 + 1 = 991\) numbers, each with 4 digits.

   Therefore, the total number of digits is:
   \[
   1 \times 9 + 2 \times 90 + 3 \times 900 + 4 \times 991 = 9 + 180 + 2700 + 3964 = 6853
   \]

2. **First step: Omit every digit on an even position:**
   - We are left with the digits at odd positions: \(\overline{A_1A_3A_5 \dots A_{6853}}\).

3. **Second step: Omit every digit on an odd position in the new sequence:**
   - We are left with the digits at positions \(3 \pmod{4}\): \(\overline{A_3A_7A_{11} \dots A_{6851}}\).

4. **Third step: Omit every digit on an even position in the new sequence:**
   - We are left with the digits at positions \(3 \pmod{8}\): \(\overline{A_3A_{11}A_{19} \dots A_{6851}}\).

5. **Fourth step: Omit every digit on an odd position in the new sequence:**
   - We are left with the digits at positions \(11 \pmod{16}\): \(\overline{A_{11}A_{27}A_{43} \dots A_{6843}}\).

6. **Fifth step: Omit every digit on an even position in the new sequence:**
   - We are left with the digits at positions \(11 \pmod{32}\).

7. **Sixth step: Omit every digit on an odd position in the new sequence:**
   - We are left with the digits at positions \(43 \pmod{64}\).

8. **Seventh step: Omit every digit on an even position in the new sequence:**
   - We are left with the digits at positions \(43 \pmod{128}\).

9. **Eighth step: Omit every digit on an odd position in the new sequence:**
   - We are left with the digits at positions \(171 \pmod{256}\).

10. **Ninth step: Omit every digit on an even position in the new sequence:**
    - We are left with the digits at positions \(171 \pmod{512}\).

11. **Tenth step: Omit every digit on an odd position in the new sequence:**
    - We are left with the digits at positions \(683 \pmod{1024}\).

12. **Eleventh step: Omit every digit on an even position in the new sequence:**
    - We are left with the digits at positions \(683 \pmod{2048}\).

13. **Twelfth step: Omit every digit on an odd position in the new sequence:**
    - We are left with the digits at positions \(2731 \pmod{4096}\).

14. **Thirteenth step: Omit every digit on an even position in the new sequence:**
    - We are left with the digits at positions \(2731 \pmod{8192}\).

    This leaves us with the digit at position \(2731\).

15. **Find the 2731st digit in the original sequence:**
    - Calculate the number of digits up to the 2731st digit:
      \[
      1 \times 9 + 2 \times 90 + 3 \times 847 + 1 = 9 + 180 + 2541 + 1 = 2731
      \]
    - The 2731st digit is the first digit of the 847th three-digit number.
    - The 847th three-digit number is \(100 + 846 = 946\).

    Therefore, the first digit of 946 is \(9\).

The final answer is \(\boxed{9}\).