II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer $N_{3}$. Erase the digits of $N_{3}$ that are in even positions, leaving the remaining digits to form the integer $N_{4}$. This process continues until only one digit remains on the blackboard. Determine this digit. (Note: Count positions from left to right, for example, in 12345, 1 is in the first position, 2 is in the second position, and so on).
$$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$

The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $\{2,4, \cdots, 6872\}$ from the set $\{1,2, \cdots, 6873\}$.

Similarly, from the remaining set $\{1,3,5, \cdots, 6873\}$, we remove $\{1,5,9, \cdots, 6873\}$, leaving $\{3,7,11, \cdots, 6871\}$. We can sequentially obtain the sets $\{3,3+8=11,11+8=19, \cdots\}$, $\{11,11+16=27, \cdots\}$, $\{11,11+32=43, \cdots\}$, $\{43,43+64=107, \cdots\}$, $\{43,43+128=371, \cdots\}$, $\{171,171+256=427, \cdots\}$, $\{171,171+512=683, \cdots\}$, $\{683,683+1024=1707, \cdots\}$, $\{683,683+2048=2731\}$, $\{2731,2731+4096=6827\}$. After the thirteenth step, only $\{2731\}$ remains.
The required digit is the 2731st digit from left to right in $N_{1}$.
$$
9 \times 1+90 \times 2<2731<9 \times 1+90 \times 2+900 \times 3 \text {, }
$$

This digit should be part of a three-digit number. The one-digit and two-digit numbers together occupy 189 positions, 2731 - 189 = 2542 = 3 \times 847 + 1, the 848th three-digit number is 947, and the digit to the left of this number is 9. The final remaining digit is 9.