$1 \cdot 200$ Write the number $1234567891011 \cdots \cdots 19941995$ on the blackboard, forming the integer $N_{1}$, erase the digits in the even positions of $N_{1}$, the remaining digits form the integer $N_{2}$, erase the digits in the odd positions of $N_{2}$, the remaining digits form the integer $N_{3}$, erase the digits in the even positions of $N_{3}$, the remaining digits form the integer $N_{4}$. This process continues until only one digit remains on the blackboard. Determine this digit (Note: Count the positions from left to right, for example, in 12345, 1 is in the 1st position, 2 is in the 2nd position, and so on).
[Solution] Since
$$
9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873,
$$

the integer $N_{1}$ consists of 6873 digits, which are numbered from left to right as 1, $2,3, \cdots, 6873$. We examine the set of these numbers $\{1,2,3, \cdots, 6873\}$.

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

Similarly, we then remove $\{1,5,9, \cdots, 6873\}$ from the remaining set $\{1,3,5, \cdots, 6873\}$, leaving the set $\{3,7,11, \cdots, 6871\}$.
Next, we can sequentially obtain the sets:
$$
\begin{array}{l}
\{3,3+8=11,11+8=19, \cdots\}, \\
\{11,11+16=27, \cdots\}, \\
\{11,11+32=43, \cdots\}, \\
\{43,43+64=107, \cdots\}, \\
\{43,43+128=171, \cdots\}, \\
\{171,171+256=427, \cdots\}, \\
\{171,171+512=683, \cdots\}, \\
\{683,683+1024=1707, \cdots\}, \\
\{683,683+2048=2731, \cdots\}, \\
\{2731,2731+4096=6827\}, \\
\{2731\} .
\end{array}
$$

Therefore, the required digit is the 2731st digit from left to right in $N_{1}$.
Since $9 \times 1+90 \times 2<2731<9 \times 1+90 \times 2+900 \times 3$, this digit belongs to a three-digit number.
Also, since $2731-9 \times 1-90 \times 2=2542$,
$$
2542=3 \times 847+1 \text {, }
$$

this digit is the first digit of the 848th three-digit number 947, which is 9.
Thus, the required digit is 9.