10.4. A magician and his assistant perform the following act: First, the assistant asks the audience to write down $N$ numbers in a row on a blackboard, then the assistant covers up two adjacent numbers. After that, the magician comes on stage and guesses the two adjacent covered numbers (including their order). To ensure that the magician can guess the result according to the prior agreement with the assistant, find the minimum value of $N$.
10.4. $N=101$.

For convenience, a sequence of $m$ digits is called an “$m$-digit number”. Suppose for some value of $N$, the magician can guess the result, so the magician can restore any two-digit number to the original $N$-digit number (the number of $N$-digit numbers that can be restored is denoted as $k_{1}$). This means that for any two-digit number, the magician can map it to an $N$-digit number (the number of such $N$-digit numbers is denoted as $k_{2}$). Therefore, $k_{1} \geqslant k_{2}$. It is easy to see that $k_{1}=(N-1) \times 10^{N-2}$ (the given two-digit number has $N-1$ ways to choose its position, and each of the remaining $N-2$ digits has 10 different ways to place the digits).
It is not difficult to see that $k_{2}=10^{N}$.
Thus, by $k_{1} \geqslant k_{2} \Rightarrow N-1 \geqslant 100 \Rightarrow N \geqslant 101$.
Below, we show that for $N=101$, the magician can guess the result.
Number the 101 digits from left to right as 0 to 100. Let the sum of the digits in all odd positions modulo 10 be $s$, and the sum of the digits in all even positions modulo 10 be $t$. Let $p=10 s+t$.

The magician and the assistant agree to cover the digits at the $p$-th and $(p+1)$-th positions. Thus, as soon as the magician sees the positions of the covered digits, he immediately knows the value of $p$. Therefore, he can determine $s$ and $t$.

Since the $p$-th and $(p+1)$-th positions are one odd and one even, once $s$ is known, the magician can calculate the digit at the covered odd position based on the sum of the digits at the uncovered odd positions.

Similarly, the magician can also calculate the digit at the covered even position.