10. We claim that if the family $\left\{A_{1}, \ldots, A_{t}\right\}$ separates the $n$-set $N$, then $2^{t} \geq n$. The proof goes by induction. The case $t=1$ is clear, so suppose that the claim holds for $t-1$. Since $A_{t}$ does not separate elements of its own or its complement, it follows that $\left\{A_{1}, \ldots, A_{t-1}\right\}$ is separating for both $A_{t}$ and $N \backslash A_{t}$, so that $\left|A_{t}\right|,\left|N \backslash A_{t}\right| \leq 2^{t-1}$. Then $|N| \leq 2 \cdot 2^{t-1}=2^{t}$, as claimed. Also, if the set $N$ with $|N|=2^{t}$ is separated by $\left\{A_{1}, \ldots, A_{t}\right\}$, then (precisely) one element of $N$ is not covered. To show this, we again use induction. This is trivial for $t=1$, so let $t \geq 1$. Since $A_{1}, \ldots, A_{t-1}$ separate both $A_{t}$ and $N \backslash A_{t}, N \backslash A_{t}$ must have exactly $2^{t-1}$ elements, and thus one of its elements is not covered by $A_{1}, \ldots, A_{t-1}$, and neither is covered by $A_{t}$. We conclude that a separating and covering family of $t$ subsets can exist only if $n \leq 2^{t}-1$. We now construct such subsets for the set $N$ if $2^{t-1} \leq n \leq 2^{t}-1, t \geq 1$. For $t=1$, put $A_{1}=\{1\}$. In the step from $t$ to $t+1$, let $N=N^{\prime} \cup N^{\prime \prime} \cup\{y\}$, where $\left|N^{\prime}\right|,\left|N^{\prime \prime}\right| \leq 2^{t-1}$; let $A_{1}^{\prime}, \ldots, A_{t}^{\prime}$ be subsets covering and separating $N^{\prime}$ and $A_{1}^{\prime \prime}, \ldots, A_{t}^{\prime \prime}$ such subsets for $N^{\prime \prime}$. Then the subsets $A_{i}=A_{i}^{\prime} \cup A_{i}^{\prime \prime}$ $(i=1, \ldots, t)$ and $A_{t+1}=N^{\prime \prime} \cup\{y\}$ obviously separate and cover $N$. The answer: $t=\left[\log _{2} n\right]+1$. Second solution. Suppose that the sets $A_{1}, \ldots, A_{t}$ cover and separate $N$. Label each element $x \in N$ with a string of $\left(x_{1} x_{2} \ldots x_{t}\right)$ of 0 's and 1's, where $x_{i}$ is 1 when $x \in A_{i}, 0$ otherwise. Since the $A_{i}$ 's separate, these strings are distinct; since they cover, the string ( $00 \ldots 0$ ) does not occur. Hence $n \leq 2^{t}-1$. Conversely, for $2^{t-1} \leq n<2^{t}$, represent the elements of $N$ in base 2 as strings of 0's and 1's of length $t$. For $1 \leq i \leq t$, take $A_{i}$ to be the set of numbers in $N$ whose binary string has a 1 in the $i$ th place. These sets clearly cover and separate.