This is one of my favourite projective geometry examples; it is a case of classification involving moduli and if done right most of the arguments can be done with a combination of geometry and linear algebra with no explicit calculation.
I assume that we are talking about an ordered quadruple of lines in $\mathbb P^3(k)$ (I do it over any field). The trick is to not think of $\mathbb P^3$ as the set of lines in $k^4$ but rather of lines in some $4$-dimensional vector space V and then use the date to get closer to an adapted coordinatisation. Assume first that no two of the lines are skew. Then a simple geometric argument shows that either all lines lie in a plane or pass through a common point. Both of those cases are reduced to the problem of four points in $\mathbb P^2$ which I skip. We can then assume that the first two lines are skew and think of $V$ as $V_1\bigoplus V_2$, where the projectivisations of $V_1$ and $V_2$ are the two first lines. Assume then that the third line is skew with the first and second line. This means that it is the projectivisation of the graph $V_3\subset V_1\bigoplus V_2$ of an isomorphism. Hence, we may assume that $V_1=V_2$ and $V_3$ is the diagonal in $V_1\bigoplus V_1$. If we also assume that the fourth line is skew with the first two, then it is also the graph $V_4\subset V_1\bigoplus V_1$ of an automorphism $V_1\rightarrow V_1$. Hence, four lines, the first two of which are skew and the last two are skew with the first two up to projective transformations correspond to isomorphism classes of pairs $(V_1,\varphi)$ where $V_1$ is a two-dimensional vector space and $\varphi$ is an automorphism of it distinct from the identity. This is the same thing as conjugacy classes of $\mathrm{GL}_2(k)$ distinct from the identity element. The condition that the last two lines be skew is exactly that $\varphi$ does not have $1$ as an eigenvalue. Of course the characteristic polynomial distinguish between conjugacy classes so there are continuous families of configuration (i.e., it has non-trivial moduli).
The remaining case of two skew lines and two lines which are not both skew with respect to both of the first lines is easy but a little bit tedious; given a pair of non skew lines one looks at the plane spanned by them and the position of the other lines with respect to it.
Addendum: I did not mean to suggest that it is the simplest problem with proper moduli. Of course the classification of four points in $\mathbb P^1$ has the cross ratio is its moduli. (The classification is proven almost word for word in the same way as the above case.)
