Determine all real-coefficient polynomials $P(x)$ such that
\[ P(\lfloor x \rfloor) = \lfloor P(x) \rfloor \]for every real numbers $x$.
1. **Assume the form of the polynomial**: Let \( P(x) = \sum_{i=0}^d a_i x^i \) be a polynomial of degree \( d \geq 1 \) with real coefficients \( a_0, a_1, \ldots, a_d \) and \( a_d \neq 0 \).

2. **Evaluate at integers**: For any integer \( n \), we have:
   \[
   \lfloor P(n) \rfloor = P(n) = \left\lfloor P\left(n + \frac{1}{2}\right) \right\rfloor.
   \]
   This implies that \( P(n) \) is an integer and:
   \[
   P(n) \leq P\left(n + \frac{1}{2}\right) < P(n) + 1.
   \]

3. **Define a new polynomial \( Q(x) \)**: Consider the polynomial \( Q(x) = P\left(x + \frac{1}{2}\right) - P(x) \). Since \( Q \) is a polynomial, we can write:
   \[
   Q(x) = \sum_{i=0}^{d-1} b_i x^i,
   \]
   where \( b_{d-1} \neq 0 \) if \( d > 1 \).

4. **Analyze \( Q(x) \) at integers**: For all integers \( n \), we have:
   \[
   0 \leq Q(n) < 1.
   \]

5. **Assume \( d > 1 \) and derive a contradiction**: Suppose \( d > 1 \). Then \( Q(x) \) has degree \( d-1 \geq 1 \). For large integers \( n \), we have:
   \[
   |Q(n)| \geq |b_{d-1}| n^{d-1} - \sum_{i=0}^{d-2} |b_i| n^i.
   \]
   To ensure \( |Q(n)| > 1 \), choose \( n \) large enough such that:
   \[
   |b_{d-1}| n > |b_{d-1}| + \max_{0 \leq i \leq d-2} |b_i| \quad \text{and} \quad |b_{d-1}| n > |b_0| + 1.
   \]
   This implies:
   \[
   |Q(n)| \geq |b_{d-1}| n - |b_0| > 1,
   \]
   which contradicts \( 0 \leq Q(n) < 1 \).

6. **Conclude \( d = 1 \)**: Since \( d > 1 \) leads to a contradiction, we must have \( d = 1 \). Thus, \( P(x) \) is a linear polynomial:
   \[
   P(x) = cx + d,
   \]
   where \( c \neq 0 \) and \( d \in \mathbb{R} \).

7. **Check the linear polynomial**: For \( P(x) = cx + d \), we need:
   \[
   P(\lfloor x \rfloor) = \lfloor P(x) \rfloor.
   \]
   This implies:
   \[
   c \lfloor x \rfloor + d = \lfloor cx + d \rfloor.
   \]
   Since \( d \) is an integer, let \( d = k \). Then:
   \[
   c \lfloor x \rfloor = \lfloor cx \rfloor.
   \]
   This holds if and only if \( c = 1 \).

8. **Final form of \( P(x) \)**: Therefore, the polynomial \( P(x) \) must be:
   \[
   P(x) = x + k,
   \]
   where \( k \) is an integer.

The final answer is \( \boxed{ P(x) = x + k } \) where \( k \) is an integer.