Problem 1638

1. **Define the set $T$ and the problem constraints:**
Given $p \geq 5$ is a prime number, and $M = \{1, 2, \ldots, p-1\}$. Define the set
$$T = \{(n, x_n) : p \mid nx_n - 1 \text{ and } n, x_n \in M\}.$$
We need to determine the value of $k$ such that
$$\sum_{(n, x_n) \in T} n \left\lfloor \frac{nx_n}{p} \right\rfloor \equiv k \pmod{p},$$
where $0 \leq k \leq p-1$ and $\left\lfloor \alpha \right\rfloor$ denotes the largest integer not exceeding $\alpha$.

2. **Identify the relationship between $n$ and $x_n$:**
Since $p \mid nx_n - 1$, we have $nx_n \equiv 1 \pmod{p}$. This implies $x_n \equiv n^{-1} \pmod{p}$, where $n^{-1}$ is the modular inverse of $n$ modulo $p$.

3. **Symmetry in the set $T$:**
Note that for each $n \in M$, there exists a unique $x_n \in M$ such that $nx_n \equiv 1 \pmod{p}$. Also, for $n \in M$, $x_{p-n} = p - x_n$.

4. Sum transformation:
Consider the sum
$$\sum_{n=1}^{p-1} n \left\lfloor \frac{nx_n}{p} \right\rfloor.$$
Since $nx_n \equiv 1 \pmod{p}$, we have $\frac{nx_n - 1}{p}$ is an integer. Thus,
$$\left\lfloor \frac{nx_n}{p} \right\rfloor = \frac{nx_n - 1}{p}.$$

5. Simplify the sum:
$$\sum_{n=1}^{p-1} n \left\lfloor \frac{nx_n}{p} \right\rfloor = \sum_{n=1}^{p-1} n \cdot \frac{nx_n - 1}{p}.$$
This can be rewritten as:
$$\sum_{n=1}^{p-1} \frac{n^2 x_n - n}{p}.$$

6. Pairing terms:
Pair the terms $n$ and $p-n$:
$$\sum_{n=1}^{p-1} \frac{n^2 x_n - n + (p-n)^2 (p-x_n) - (p-n)}{p}.$$
Simplify the expression inside the sum:
$$\frac{n^2 x_n - n + (p-n)^2 (p-x_n) - (p-n)}{p} = \frac{n^2 x_n - n + (p^2 - 2pn + n^2)(p - x_n) - (p-n)}{p}.$$

7. Further simplification:
$$= \frac{n^2 x_n - n + p^3 - p^2 x_n - 2p^2 n + 2pn x_n + n^2 p - p - p + n}{p}.$$
Combine like terms:
$$= \frac{p^3 - p^2 x_n - 2p^2 n + 2pn x_n + n^2 p - p}{p}.$$

8. Final simplification:
$$= p^2 - px_n - 2pn + 2nx_n + n^2 - 1.$$
Summing over $n$ from 1 to $p-1$:
$$\sum_{n=1}^{p-1} (p^2 - px_n - 2pn + 2nx_n + n^2 - 1).$$

9. **Modulo $p$ simplification:**
Since $p^2 \equiv 0 \pmod{p}$ and $px_n \equiv 0 \pmod{p}$:
$$\sum_{n=1}^{p-1} (2 + n^2 - 1) = \sum_{n=1}^{p-1} (n^2 + 1).$$

10. Sum of squares and constants:
$$\sum_{n=1}^{p-1} n^2 = \frac{(p-1)p(2p-1)}{6},$$
and
$$\sum_{n=1}^{p-1} 1 = p-1.$$

11. Combine results:
$$\sum_{n=1}^{p-1} (n^2 + 1) = \frac{(p-1)p(2p-1)}{6} + (p-1).$$

12. **Final modulo $p$ calculation:**
$$\frac{(p-1)p(2p-1)}{6} + (p-1) \equiv \frac{(p-1)(2p-1)}{6} + (p-1) \pmod{p}.$$
Simplify:
$$\frac{(p-1)(2p-1 + 6)}{6} = \frac{(p-1)(2p+5)}{6}.$$
Since $2p+5 \equiv 2p+5 \pmod{p}$:
$$\frac{(p-1)(2p+5)}{6} \equiv \frac{(p-1)(2p+5)}{6} \pmod{p}.$$

13. Final result:
$$k = \frac{p-1}{2}.$$

The final answer is $\boxed{\frac{p-1}{2}}$.