Given a permutation $(a_{1}, a_{1},...,a_{n})$ of the numbers $1, 2,...,n$ one may interchange any two consecutive "blocks" - that is, one may transform 
($a_{1}, a_{2},...,a_{i}$,$\underbrace {a_{i+1},... a_{i+p},}_{A} $ $ \underbrace{a_{i+p+1},...,a_{i+q},}_{B}...,a_{n}) $
into 
$ (a_{1}, a_{2},...,a_{i},$ $ \underbrace {a_{i+p+1},...,a_{i+q},}_{B} $ $ \underbrace {a_{i+1},... a_{i+p}}_{A}$$,...,a_{n}) $
by interchanging the "blocks" $A$ and $B$. Find the least number of such changes which are needed to transform $(n, n-1,...,1)$ into $(1,2,...,n)$
To solve the problem, we need to determine the minimum number of block interchanges required to transform the permutation $(n, n-1, \ldots, 1)$ into $(1, 2, \ldots, n)$. We will use induction to prove the result.

1. **Base Case:**
   - For \( n = 2 \):
     - The permutation is \((2, 1)\).
     - We can swap the blocks \((2)\) and \((1)\) to get \((1, 2)\).
     - This requires 1 move.
     - Therefore, the base case holds with 1 move.

2. **Inductive Step:**
   - Assume the result holds for \( n = k \), i.e., it takes \( \left\lfloor \frac{k+1}{2} \right\rfloor \) moves to transform \((k, k-1, \ldots, 1)\) into \((1, 2, \ldots, k)\).
   - We need to show that the result holds for \( n = k+1 \).

3. **Transforming \( (k+1, k, \ldots, 1) \) to \( (1, 2, \ldots, k+1) \):**
   - Consider the permutation \((k+1, k, \ldots, 1)\).
   - We can swap the block \((k+1)\) with the block \((k, k-1, \ldots, 1)\) to get \((k, k-1, \ldots, 1, k+1)\).
   - Now, we need to transform \((k, k-1, \ldots, 1)\) into \((1, 2, \ldots, k)\), which by the inductive hypothesis takes \( \left\lfloor \frac{k+1}{2} \right\rfloor \) moves.

4. **Combining the Moves:**
   - The initial swap of \((k+1)\) with \((k, k-1, \ldots, 1)\) takes 1 move.
   - Transforming \((k, k-1, \ldots, 1)\) into \((1, 2, \ldots, k)\) takes \( \left\lfloor \frac{k+1}{2} \right\rfloor \) moves.
   - Therefore, the total number of moves required is \( 1 + \left\lfloor \frac{k+1}{2} \right\rfloor \).

5. **Conclusion:**
   - For \( n = k+1 \), the number of moves required is \( \left\lfloor \frac{(k+1)+1}{2} \right\rfloor = \left\lfloor \frac{k+2}{2} \right\rfloor \).
   - This matches the form \( \left\lfloor \frac{n+1}{2} \right\rfloor \).

Thus, by induction, the least number of block interchanges needed to transform \((n, n-1, \ldots, 1)\) into \((1, 2, \ldots, n)\) is \( \left\lfloor \frac{n+1}{2} \right\rfloor \).

The final answer is \( \boxed{ \left\lfloor \frac{n+1}{2} \right\rfloor } \).