There are $2n$ different numbers in a row. By one move we can interchange any two numbers or interchange any $3$ numbers cyclically (choose $a,b,c$ and place $a$ instead of $b$, $b$ instead of $c$, $c$ instead of $a$). What is the minimal number of moves that is always sufficient to arrange the numbers in increasing order ?
To determine the minimal number of moves required to arrange $2n$ different numbers in increasing order, we need to consider the types of moves allowed: interchanging any two numbers or cyclically interchanging any three numbers. Let's break down the solution step-by-step.

1. **Proving that at least two numbers can be placed correctly in one move:**
   - By interchanging any two numbers, we can place at least one number in its correct position.
   - By cyclically interchanging any three numbers, we can place at least one number in its correct position.
   - Therefore, in each move, we can place at least one number correctly. However, in some cases, we can place two numbers correctly in one move.

2. **Considering the worst-case scenario:**
   - Consider the permutation $(2n, 2n-1, n+1, n, \ldots, 2, 1)$.
   - Divide the row into two parts: the first $n$ numbers and the second $n$ numbers.
   - In this permutation, all $2n$ numbers are not in the parts they are supposed to be.
   - Each move can place at most two numbers into their appropriate parts.

3. **Calculating the minimal number of moves:**
   - Since each move can place at most two numbers correctly, and there are $2n$ numbers to be placed correctly, we need at least $\frac{2n}{2} = n$ moves.
   - Therefore, the minimal number of moves required to arrange the numbers in increasing order is $n$.

The final answer is $\boxed{n}$.