Let $n \ge 2$ be an integer and $f_1(x), f_2(x), \ldots, f_{n}(x)$ a sequence of polynomials with integer coefficients.  One is allowed to make moves $M_1, M_2, \ldots $ as follows: in the $k$-th move $M_k$ one chooses an element $f(x)$ of the sequence  with degree of $f$ at least $2$ and replaces it with $(f(x) - f(k))/(x-k)$.  The process stops when all the elements of the sequence are of degree $1$.  If $f_1(x) = f_2(x) = \cdots = f_n(x) = x^n + 1$, determine whether or not it is possible to make appropriate moves such that  the process stops with a sequence of $n$ identical polynomials of degree 1.
1. **Lemma: Suppose that \( f(x) \) is a monic polynomial. Then, the move \( M_k \) increases the vice-leading coefficient by \( k \).**

   **Proof of Lemma:**
   Let \( f(x) = \sum_{i=0}^n a_i x^i \), where \( a_n = 1 \). Therefore,
   \[
   \frac{f(x) - f(k)}{x - k} = \frac{\sum_{i=0}^n a_i x^i - \sum_{i=0}^n a_i k^i}{x - k}
   \]
   By polynomial division, this results in:
   \[
   \frac{f(x) - f(k)}{x - k} = x^{n-1} + (k + a_{n-1}) x^{n-2} + \cdots
   \]
   Here, the term \( k + a_{n-1} \) is the new vice-leading coefficient. Thus, the move \( M_k \) increases the vice-leading coefficient by \( k \).

2. **Using the Lemma:**
   Each move reduces the degree of the polynomial by 1 while increasing the vice-leading coefficient by the index of the move. To reduce each polynomial \( f_i(x) \) from degree \( n \) to degree 1, we need \( n-1 \) moves per polynomial. Since there are \( n \) polynomials, a total of \( n(n-1) \) moves are required.

3. **Ensuring Identical Linear Polynomials:**
   To ensure that all resulting polynomials are identical linear polynomials, the sum of the indices of the moves made on each polynomial must be equal. This is because the vice-leading coefficient must be the same for all polynomials after all moves are completed.

4. **Sum of Indices:**
   The sum of the indices of the moves made on each polynomial must be equal. The total sum of indices for all moves is:
   \[
   1 + 2 + \cdots + n(n-1) = \frac{n(n-1)(n^2 - n + 1)}{2}
   \]
   For this sum to be divisible by \( n \), \( n \) must be odd. This is because the sum of the first \( n(n-1) \) natural numbers is divisible by \( n \) if and only if \( n \) is odd.

5. **Construction for Odd \( n \):**
   For odd \( n \), we can construct the sequence of moves as follows:
   - Apply the moves \( M_{2kn+i} \) and \( M_{2(k+1)n+1-i} \) on the polynomial \( f_i(x) \) for all \( k \) from 0 to \( \frac{n-3}{2} \).
   - This ensures that all moves are covered exactly once and the vice-leading coefficients remain equal after each pair of moves.

6. **Conclusion:**
   Since the construction works for all odd \( n \), it is possible to make appropriate moves such that the process stops with a sequence of \( n \) identical polynomials of degree 1.

The final answer is that it is possible for all odd \( \boxed{ n } \).