For a polynomial $ P$ of degree 2000 with distinct real coefficients let $ M(P)$ be the set of all polynomials that can be produced from $ P$ by permutation of its coefficients. A polynomial $ P$ will be called [b]$ n$-independent[/b] if $ P(n) \equal{} 0$ and we can get from any $ Q \in M(P)$ a polynomial $ Q_1$ such that $ Q_1(n) \equal{} 0$ by interchanging at most one pair of coefficients of $ Q.$ Find all integers $ n$ for which $ n$-independent polynomials exist.
1. **Remove the condition \( P(n) = 0 \)**:
   The condition \( P(n) = 0 \) is superfluous because we are interested in the properties of the polynomial's coefficients and their permutations. Thus, we can ignore this condition for now.

2. **Define the problem for \( d = 2000 \)**:
   Let \( d = 2000 \). We need to find all integers \( n \) for which \( n \)-independent polynomials of degree \( d \) exist.

3. **Case 1: \( |n| \ge 2 \)**:
   - **Good and Bad Permutations**:
     A permutation \( \pi \) is called *good* if \( \sum_{k=0}^{d} a_{\pi(k)} n^k = 0 \), and *bad* otherwise.
   
   - **Rational Approximation**:
     By rational approximation, we can find pairwise distinct integers \( b_0 < b_1 < \cdots < b_d \) such that \( \sum b_{\pi(k)} n^k = 0 \) if and only if \( \sum a_{\pi(k)} n^k = 0 \). This follows from the \( d+1 \)-dimensional Kronecker's theorem.

   - **Lemma**:
     Suppose there are \( r_i \) indices \( j \) with \( b_j \equiv i \pmod{n} \). Then \( \sum_{k=0}^{d} b_{\pi(k)} n^k = C \) has at most \( r_0! \cdots r_{n-1}! \) solutions \( \pi \). Furthermore, if equality holds, then for each \( j \) with \( b_j \equiv C \pmod{n} \), \( \sum_{k=0}^{d-1} b_{\sigma(k)} n^k = \frac{C - b_j}{n} \) must have exactly \( r_0! \cdots (r_C - 1)! \cdots r_{n-1}! \) solutions \( \sigma \) (with \( \sigma \) restricted to the indices \( k \ne j \)); in particular, there must be at least one index \( k \ne j \) with \( b_k \equiv \frac{C - b_j}{n} \pmod{n} \).

   - **Proof of Lemma**:
     Induct on \( r_0 + \cdots + r_{n-1} = d+1 \ge 1 \) using the requirement \( a_{\pi(0)} \equiv C \pmod{n} \). \( \blacksquare \)

   - **Contradiction**:
     Assume \( \gcd(b_0, b_1, \ldots, b_d) = 1 \), so \( 0 < r_0 < d+1 \), and there exists \( 1 \le t \le n-1 \) with \( r_t > 0 \); WLOG \( b_d \equiv t \pmod{n} \). Now focus on the \( \pi \) with \( \pi(0) = d \): they must all be bad. On the other hand, any good \( \pi \) "saves" (via a single transposition) exactly one bad \( \pi' \) with \( \pi'(0) = d \). But each bad \( \pi' \) (with \( \pi'(0) = d \)) is saved by at least one good \( \pi \), so we must have at least \( d! \) good \( \pi \). Yet by the lemma and the fact that \( a! b! < (a+1)! (b-1)! \) whenever \( a \ge b \ge 1 \), we have at most \( d! 1! \) good \( \pi \), with equality only when \( \{r_0, r_t\} = \{1, d\} \).

     If \( r_0 = 1 \), then WLOG \( b_0 \equiv 0 \pmod{n} \) and \( b_1, \ldots, b_d \equiv t \pmod{n} \). It follows that \( \sum_{k=0}^{d-1} b_{\pi(k)} n^k = \frac{-b_0}{n} \) for all \( \pi \), from which we easily obtain \( b_1 = \cdots = b_d \) (e.g., consider the effect of a transposition), contradiction.

     Otherwise, \( r_t = 1 \), so \( b_0, \ldots, b_{d-1} \equiv 0 \pmod{n} \). Take a good \( \pi \) and suppose \( \pi(M) = d \). If \( M \le d-2 \), let \( \{u, v\} = \{0, 1, \ldots, d\} \setminus \{\pi(0), \ldots, \pi(d-2)\} \). Then by repeated application of the equality clause, \( b_{\sigma(d-1)} + n b_{\sigma(d)} = -\frac{\sum_{k=0}^{d-2} b_{\pi(k)} n^k}{n^{d-1}} \) has \( 2! \) solutions \( \sigma \) (where \( \{\sigma(d-1), \sigma(d)\} = \{u, v\} \)), since \( b_u, b_v \equiv 0 \pmod{n} \); hence \( b_u = b_v \), which is impossible.

     Thus \( M \ge d-1 \) for any \( \pi \), whence the equality clause gives us \( n^k \mid b_0, \ldots, b_{d-1} \) for \( k = 1, 2, \ldots, d-2 \) (just induct on \( k \)). Now fix some \( \pi \); the equality clause tells us there exists \( \pi' \ne \pi \) with \( \pi'(i) = \pi(i) \) for \( i = 0, 1, \ldots, d-3 \). We cannot have \( \pi'(d-2) = d \) (by the previous paragraph) or \( \pi'(M) = \pi(M) = d \) (\( \pi', \pi \) can't agree in any more places), so WLOG \( M = d-1 \implies \pi(d-1) = d \) and \( \pi'(d) = d \). Thus
     \[
     0 = \sum (b_{\pi(k)} - b_{\pi'(k)}) n^k = (b_{\pi(d-2)} - b_{\pi'(d-2)}) n^{d-2} + (b_d - b_{\pi'(d-1)}) n^{d-1} + (b_{\pi(d)} - b_d) n^d,
     \]
     which is impossible mod \( n^d \) (we have \( n^2 \mid n^{d-2} \mid b_{\pi(d-2)}, b_{\pi'(d-2)}, b_{\pi'(d-1)} \), but \( n \nmid b_d \)).

4. **Case 2: \( n = -1 \)**:
   - For any indices \( i_1 < i_2 < \cdots < i_{1000} \), there exist indices \( j, k \) with
     \[
     a_{i_1} + \cdots + a_{i_{1000}} = C + (a_{i_j} - a_k),
     \]
     where \( 2C = a_0 + \cdots + a_{2000} \). This already tells us that
     \[
     (a_{1001} + \cdots + a_{2000}) - (a_0 + \cdots + a_{999}) \le 2(a_{2000} - a_0),
     \]
     which although "unlikely," still needs a small improvement. Now note that \( \{a_0 + a_1 + a_{i_3} + \cdots + a_{i_{1000}}\}_{1 < i_3 < \cdots < i_{1000} < 1999} \) has at least \( 1 + 998 \cdot 999 \) elements, one of which must, by pigeonhole, take the form \( C + a_u - a_v \) for \( u, v \ne 0, 1, 1999, 2000 \). But for this choice of \( (i_j) \), we then have
     \[
     (a_{i_3} + \cdots + a_{i_{1000}} + a_{1999} + a_{2000}) - (a_0 + a_1 + a_{i_3} + \cdots + a_{i_{1000}}) \le (C + a_{2000} - a_0) - (C + a_u - a_v) \le (a_{2000} - a_0) + (a_{1998} - a_2),
     \]
     which is absurd.

Conclusion:
The only possible values for \( n \) are \( 0 \) and \( 1 \).

The final answer is \( \boxed{ n = 0 } \) or \( n = 1 \).