Let $n$ be a natural number. Two numbers  are called  "unsociable" if their greatest common divisor is $1$. The numbers $\{1,2,...,2n\}$ are partitioned into $n$ pairs. What is the minimum number of "unsociable" pairs that are formed?
1. **Understanding the Problem:**
   We need to partition the set $\{1, 2, \ldots, 2n\}$ into $n$ pairs such that the number of "unsociable" pairs (pairs whose greatest common divisor is 1) is minimized.

2. **Identifying Key Elements:**
   - A pair $(a, b)$ is "unsociable" if $\gcd(a, b) = 1$.
   - We need to find the minimum number of such pairs.

3. **Lower Bound Analysis:**
   - Consider the primes between $n+1$ and $2n$. Each of these primes will form an unsociable pair because a prime number greater than $n$ cannot share any common divisor with any number less than or equal to $n$ except 1.
   - Additionally, the number 1 will always form an unsociable pair with any other number.

4. **Counting the Relevant Primes:**
   - Let $\pi(x)$ denote the number of primes less than or equal to $x$.
   - The number of primes between $n+1$ and $2n$ is $\pi(2n) - \pi(n)$.
   - Including the number 1, we have $\pi(2n) - \pi(n) + 1$ numbers that will form unsociable pairs.

5. **Formulating the Minimum Number of Unsociable Pairs:**
   - Since each pair consists of two numbers, the minimum number of unsociable pairs is given by:
     \[
     x = \left\lceil \frac{\pi(2n) - \pi(n) + 1}{2} \right\rceil
     \]

6. **Constructing the Pairs:**
   - Let $S_0 \subseteq \{1, 2, \ldots, 2n\}$ be the set of numbers excluding 1 and the primes between $n+1$ and $2n$.
   - We need to pair the elements of $S_0$ such that $\gcd(i, j) > 1$ for each pair $(i, j)$.

7. **Algorithm for Pairing:**
   - Let $p_1 < p_2 < \ldots < p_k$ be the primes in $S_0$.
   - Define $X = \emptyset$.
   - For each step $0 \le i \le k-1$:
     - Define $T_i$ as the set of numbers in $S_i$ divisible by $p_{k-i}$.
     - If $|T_i|$ is even, pair the elements arbitrarily and remove them from $S_i$.
     - If $|T_i|$ is odd, pair all elements except $2p_{k-i}$ and remove them from $S_i$. Add $2p_{k-i}$ to $X$ if it is not paired.
   - After step $k-1$, the only unpaired numbers are in $X$.
   - Since all elements in $X$ are of the form $2p_i$ (even numbers), pair them arbitrarily until at most one number is left.

8. **Conclusion:**
   - The minimum number of unsociable pairs is given by:
     \[
     x = \left\lceil \frac{\pi(2n) - \pi(n) + 1}{2} \right\rceil
     \]

The final answer is \( \boxed{ \left\lceil \frac{\pi(2n) - \pi(n) + 1}{2} \right\rceil } \)