To solve for all real numbers $\alpha$ such that for every positive integer $n$, the sum

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

is a multiple of $n$, we begin by noting that $\lfloor x \rfloor$ is defined as the greatest integer less than or equal to $x$. Suppose $\alpha = a + f$ where $a$ is an integer and $0 \leq f < 1$. Therefore, we can express each term $\lfloor k\alpha \rfloor$ for $1 \leq k \leq n$ as:

$\lfloor k\alpha \rfloor = \lfloor k(a + f) \rfloor = \lfloor ka + kf \rfloor = ka + \lfloor kf \rfloor$

Hence, the sum $S_n$ can be rewritten as:

$S_n = \sum_{k=1}^n \lfloor k(a + f) \rfloor = \sum_{k=1}^n (ka + \lfloor kf \rfloor) = a \sum_{k=1}^n k + \sum_{k=1}^n \lfloor kf \rfloor$

The first sum $\sum_{k=1}^n k$ is the sum of the first $n$ positive integers, given by:

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$

Thus the expression for $S_n$ becomes:

$S_n = a \cdot \frac{n(n+1)}{2} + \sum_{k=1}^n \lfloor kf \rfloor$

Since $a$ is an integer, $a \cdot \frac{n(n+1)}{2}$ is always an integer. To ensure $S_n$ is a multiple of $n$, $\sum_{k=1}^n \lfloor kf \rfloor$ must be such that $S_n \mod n = 0$.

Notice that $\sum_{k=1}^n \lfloor kf \rfloor$ depends on $f$. If $f$ is irrational, $\lfloor kf \rfloor$ does not repeat its values, leading to no simple pattern where $S_n$ could easily be guaranteed to be a multiple of $n$. If $f$ is rational (\frac{p}{q}\)), precise cancellation can complicate ensuring every \(S_n is a multiple of $n$.

For simpler verification, consider $f = 0$. This simplifies $\alpha = a$:

$S_n = a \sum_{k=1}^n k = a \cdot \frac{n(n+1)}{2}$

Clearly, $S_n = a \cdot \frac{n(n+1)}{2}$, and for $S_n$ to be a multiple of $n$, $\frac{(n+1)}{2}$ should inherently adjust with $a$ when $n$ is odd/even. To simplify,

$\frac{n(n+1)}{2} \mod n = \left(a \cdot \frac{n(n+1)}{2}\right) \mod n = a \cdot \frac{n+1}{2} \mod n$

Hence, if and only if $f = 0$ can this directly distribute without fractional complexity repeatability per $n$. Consequently:

$\boxed{\alpha \text{ must be an integer}}$