$10 \cdot 114$ Write the number $123456789101112 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits in the even positions of $N_{1}$, leaving the remaining digits to form the integer $N_{2}$. Erase the digits in the odd positions of $N_{2}$, leaving the remaining digits to form the integer $N_{3}$. Erase the digits in the even positions of $N_{3}$, 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)
(Australian Mathematical Olympiad, 1995)
[Solution] The integer $N_{1}$ is composed of $9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873$ digits. We number these digits as $1,2,3, \cdots, 6873$, and consider the set $\{1,2,3, \cdots, 6873\}$.

Erasing the digits at even positions from $N_{1}$ is equivalent to removing $\{2,4, \cdots, 6872\}$ from the set $\{1,2, \cdots, 6874\}$, leaving the set $\{1,3,5, \cdots, 6873\}=\{1,1+2=3,3+2=5, \cdots, 6871+2=6873\}$,

Next, from the set $\{1,3,5, \cdots, 6873\}$, remove $\{1,5,9, \cdots, 6873\}$, resulting in the remaining set $\{3,7,11, \cdots, 6781\}=\{3,3+4=7,7+4=11, \cdots, 6867+4=6871\}$.

Then, from the remaining set $\{3,7,11, \cdots, 6871\}$, remove $\{7,15, \cdots\}$, resulting in the set $\{3,11, \cdots\}=\{3,3+8=11,11+8=19, \cdots\}$,
and so on, we get
$$
\begin{array}{l}
\quad\{11,11+16=27, \cdots\},\{11,11+32=43, \cdots\},\{43,43+64=107, \\
\cdots\{,\{43,43+128=171, \cdots\},\{171,171+256=427, \cdots\},\{171,171+ \\
512=683, \cdots\},\{683,683+1024=1707, \cdots\},\{683,683+2048=2731\}, \\
\{2731,2731+4096=6827\},
\end{array}
$$

After the thirteenth step, only $\{2731\}$ remains.
The required digit is the 2731st digit from left to right in $N$. Since $9 \times 1+90$ $\times 2<2731<9 \times 1+90 \times 2+900 \times 3$,
the number is a three-digit number.
The one-digit and two-digit numbers occupy $9 \times 1+90 \times 2=189$ positions, and $2731-189$ $=2542=3 \times 847+1$, so, the 848th number in the three-digit numbers is 947, and its first digit is 9. Therefore, the last remaining digit on the blackboard is 9.