10. Answer: 99 .
First assume that $n \geq 100$. Consider the following set $A$ :
$$
A=\{1+11 i: i=0,1,2, \cdots, 9\} .
$$

Note that $|A|=10, A \subseteq N_{n}$ and

Hence the answer is not larger than 99 .
Now consider the case that $n \leq 99$. For $i=1,2,3, \cdots, 9$,
$$
P_{1}=\{11(i-1)+j: 1 \leq j \leq 11\} .
$$

Note that
$$
N_{n} \subseteq \bigcup_{1 \leq i \leq 9} P_{i}
$$

Now assume that $A$ is any subset of $N_{n}$ with $|A| \geq 10$. By Pigeonhole Principle, $\left|A \cap P_{i}\right| \geq$ 2 for some $i: 1 \leq i \leq 9$. Let $c_{1}, c_{2} \in A \cap P_{1}$. Thus
$$
\underset{\substack{a . b \in A \\ a \neq 4}}{\mathrm{i}}|a-b| \leq\left|c_{1}-c_{2}\right| \leq(11(i-1)+11)-(11(i-1)+1)=10 \text {. }
$$
10. For any positive integer $n$, let $N_{n}$ be the set of integers from 1 to $n$, i.e., $N_{n}=$ $\{1,2,3, \cdots, n\}$. Now assume that $n \geq 10$. Determine the maximum value of $n$ such that the following inequality
$$
\mathrm{i}_{\substack{a, b \in A \\ a \neq b}}|a-b| \leq 10
$$
holds for each $A \subseteq N_{n}$ with $|A| \geq 10$.