To solve the problem, we need to determine all real numbers $\alpha$ such that, for every positive integer $n$, the integer

$\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor$

is a multiple of $n$. Here, $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

Step-by-Step Solution:

1. Express the Sum: Consider the sum $S_n = \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor$.

2. Fractional Part Decomposition: Let $\alpha = k + f$ where $k$ is an integer and $0 \leq f < 1$ (i.e., $f$ is the fractional part of $\alpha$).

3. Analyze Each Term: Each term $\lfloor j\alpha \rfloor$ can be written as: $\lfloor j\alpha \rfloor = \lfloor j(k + f) \rfloor = \lfloor jk + jf \rfloor$ Since $k$ is an integer: $\lfloor jk + jf \rfloor = jk + \lfloor jf \rfloor$ So, we have: $\lfloor j\alpha \rfloor = jk + \lfloor jf \rfloor$

4. Sum the Terms: The sum $S_n$ becomes: $S_n = \sum_{j=1}^{n} (jk + \lfloor jf \rfloor) = k\sum_{j=1}^{n} j + \sum_{j=1}^{n} \lfloor jf \rfloor$ The first part of the sum is: $k \sum_{j=1}^{n} j = k \cdot \frac{n(n+1)}{2}$

5. Multiples of $n$: We need $S_n$ to be a multiple of $n$: $S_n = k \cdot \frac{n(n+1)}{2} + \sum_{j=1}^{n} \lfloor jf \rfloor$ must be a multiple of $n$. The first term $k \cdot \frac{n(n+1)}{2}$ is generally not a multiple of $n$ unless $k = 0$.

6. Simplify: If $k = 0$, $\alpha = f$, and $S_n = \sum_{j=1}^{n} \lfloor jf \rfloor$ must be a multiple of $n$.

7. Periodic Behavior of the Fractional Part: For $S_n = \sum_{j=1}^{n} \lfloor jf \rfloor$ to be a multiple of $n$, $f$ must be such that the sequence $\lfloor jf \rfloor$ behaves in a periodic manner or exhibits a specific structure.

8. Special Values: If $f$ is rational, say $f = \frac{p}{q}$ in lowest terms, the sequence $\lfloor jf \rfloor$ will have a periodic structure with period $q$. However, the sum over a period is not generally guaranteed to be a multiple of $n$.

9. Conclusion: Given the requirements, the only viable solution is $f = 0$, meaning $\alpha$ must be an integer.

Therefore, the only real numbers $\alpha$ such that the given condition holds are the integers:

$\boxed{\alpha \in \mathbb{Z}}$

Would you like further details or have any questions?

Here are some questions you can ask:

1. How do you generalize the solution for non-integer $\alpha$?
2. Can you show an example with a specific $\alpha$ value?
3. Why can't non-integer values of $\alpha$ work?
4. How does the behavior of fractional parts affect the sum?
5. Can you prove this using a different method?

Tip: When dealing with floor functions and sums, decomposing into integer and fractional parts can greatly simplify the problem.