(11. $\rightarrow$ A magician and his assistant perform the following act: First, the assistant asks the audience to write $N$ numbers in a row on the blackboard; then the assistant covers two adjacent numbers; subsequently, the magician comes on stage and guesses the two adjacent covered numbers (including their order). To ensure the magician can guess the result according to the prior agreement with the assistant, find the minimum value of $N$.
Solve $N=101$.
For convenience, we will call a sequence of $m$ numbers an “$m$-digit number”. Suppose for some value of $N$, the magician can guess the result. Thus, 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 $k_{1}=(N-1) \cdot 10^{N-2}$ (the given “two-digit number” has $N-1$ ways to choose positions, and each of the remaining $N-2$ positions can hold 10 different digits). It is also not difficult to see that $k_{2}=10^{N}$. Thus, from $k_{1} \geqslant k_{2}$, we derive $N-1 \geqslant 100$, i.e., $N \geqslant 101$.

Next, we will show that for $N=101$, the magician can guess the result. Number the 101 positions from left to right as 0 to 100. Let the sum of the digits in all odd-numbered positions modulo 10 be $s$, and the sum of the digits in all even-numbered positions modulo 10 be $t$. Let $p=10 s+t$. The magician and the assistant agree to cover the digits in 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 even and one odd, knowing $s$ allows him to calculate the digit in the covered odd position based on the sum of the digits in the uncovered odd positions. Similarly, he can calculate the digit in the covered even position.