Ukrainian National Mathematical Olympiad

First, we go through cases when this equation is not quadratic.
For $k=0$, equation becomes $-4x+6=0$, and has a non-integer solution.
For $k=2$, equation becomes $-6x+6=0$, and has a single solution $x=1$, which satisfies the statement.

Now, suppose $k(2-k) \neq 0$. We find the discriminant of this quadratic equation:
$$D = (k^2 + 8k + 16) - 24(2k - k^2) = 25k^2 - 40k + 16 = (5k - 4)^2$$
So the roots are:
$$x_1, x_2 = \frac{(k+4) \pm (5k-4)}{2k(2-k)}$$
Calculating both roots:
$$x_1 = \frac{(k+4) + (5k-4)}{2k(2-k)} = \frac{6k}{2k(2-k)} = \frac{3}{2-k}$$
$$x_2 = \frac{(k+4) - (5k-4)}{2k(2-k)} = \frac{-4k+8}{2k(2-k)} = \frac{2}{k}$$
We must find all cases when this is a positive integer. Let $\frac{2}{k} = m$ be a positive integer, then $k = \frac{2}{m}$.

Then:
$$\frac{3}{2-k} = \frac{3}{2-\frac{2}{m}} = \frac{3m}{2(m-1)}$$
We require $\frac{3m}{2(m-1)}$ to be a positive integer. Since $m$ is a positive integer, $m$ and $m-1$ are coprime.
Hence, the last condition is only possible for $m-1=1 \Rightarrow k=1$ or $m-1=3 \Rightarrow k=\frac{1}{2}$.

Therefore, the real values of $k$ are $k=2$, $k=1$, and $k=\frac{1}{2}$.