Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$. Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by the prime $2027$.
Given the monic polynomial \( P(x) \) of degree 2023 such that \( P(k) = k^{2023} P\left(1 - \frac{1}{k}\right) \) for every positive integer \( 1 \leq k \leq 2023 \), we need to find \( P(-1) \) in the form \( \frac{a}{b} \) where \( a \) and \( b \) are relatively prime integers. Then, we need to compute the unique integer \( 0 \leq n < 2027 \) such that \( bn - a \) is divisible by the prime 2027.

1. **Understanding the Functional Equation:**
   \[
   P(k) = k^{2023} P\left(1 - \frac{1}{k}\right)
   \]
   This implies that if \( k \) is a root of \( P(x) \), then \( 1 - \frac{1}{k} \) is also a root. 

2. **Analyzing the Polynomial:**
   Since \( P(x) \) is monic and of degree 2023, we can write:
   \[
   P(x) = x^{2023} + a_{2022} x^{2022} + \cdots + a_1 x + a_0
   \]

3. **Using the Functional Equation:**
   Substitute \( k = 1 \):
   \[
   P(1) = 1^{2023} P(0) = P(0)
   \]
   This implies \( P(1) = P(0) \).

4. **Substitute \( k = 2 \):**
   \[
   P(2) = 2^{2023} P\left(1 - \frac{1}{2}\right) = 2^{2023} P\left(\frac{1}{2}\right)
   \]

5. **Substitute \( k = -1 \):**
   \[
   P(-1) = (-1)^{2023} P\left(1 - \frac{1}{-1}\right) = -P(2)
   \]
   Since \( P(-1) = -P(2) \), we need to find \( P(2) \).

6. **Finding \( P(2) \):**
   From the previous step:
   \[
   P(2) = 2^{2023} P\left(\frac{1}{2}\right)
   \]
   Substitute \( k = \frac{1}{2} \):
   \[
   P\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{2023} P\left(1 - 2\right) = \left(\frac{1}{2}\right)^{2023} P(-1)
   \]
   Therefore:
   \[
   P(2) = 2^{2023} \left(\frac{1}{2}\right)^{2023} P(-1) = P(-1)
   \]
   This implies:
   \[
   P(-1) = -P(2) = -P(-1)
   \]
   Hence:
   \[
   2P(-1) = 0 \implies P(-1) = 0
   \]

7. **Conclusion:**
   Since \( P(-1) = 0 \), we have \( a = 0 \) and \( b = 1 \). We need to find \( n \) such that \( bn - a \) is divisible by 2027:
   \[
   1n - 0 \equiv 0 \pmod{2027}
   \]
   Therefore:
   \[
   n \equiv 0 \pmod{2027}
   \]

The final answer is \( \boxed{0} \).