If the Sombor index of a graph $G$ is integer, then the vertices of $G$ of the same degree form a stable (aka independent) set, and $G$ has minimum degree at least three. This essentially follows from arguments similar to the answer here. So we can write $V(G)$ has a disjoint union of sets of form $V_i$ where $V_i$ consists of vertices of degree $i$. What can we say about the edges of $G$? As previously observed, no edges joint two vertices of $V_i$, so the question becomes for what $i$ and $j$ can there be edges between $V_i$ and $V_j$? This is answered in the linked answer above: namely, such $i$ and $j$ must be the two smallest numbers of some Pythagorean triple $(i, j, k)$. Let us call such $i, j$ a Pythagorean pair. The previous description is a complete characterization, meaning that $G$ has integer Sombor index if and only if $G$ has the described structure above.
We now give some necessary conditions for a line graph to have integer Sombor index (if it exists). If $SO(L(G))$ is an integer, $L(G)$ must have the above structure. The question is then are there line graphs with the previous structure? Beineke gave an induced subgraph characterization of line graphs in terms of 9 obstructions. This means that $G$ is the line graph of some other graph if and only if $G$ does not contain one of these 9 obstructions as induced subgraphs.
The simplest of the obstructions is $K_{1,3}$. Thus, among any three neighbours of any vertex $v$, there must be an edge. In particular, this implies that $v$ is in a triangle (a complete graph on 3 vertices). In fact, it is in many triangles (one for each triple of neighbours). So if $v\in V_i$ then there exists $u\in V_j$ and $w\in V_k$ such that $i, j$, and $i, k$ and $j, k$ all form Pythagorean pairs. Especially, they come from an Euler brick. Assuming $i\leq j\leq k$, the size of a smallest Euler brick (see previous link) implies that $i\geq 44, j\geq 117, k\geq 240$. So, $G$ has minimum degree $44$. But we said that among any three neighbours of $v$, there must be an edge. Thus, at most two neighbours of $v$ can come from the same $V_j$, i.e., at most two neighbours of $v$ have the same degree. Since the degree of each neighbour of $v$ must be able to form an Euler brick with the degree of $v$, the degree of $v$ must be in many Euler bricks with distinct parameters. These exist (see my question), but the numbers get big quickly (think $>>10^{10}$). And we've only thought about one vertex, but the above must be true for every vertex. And we have yet to take into consideration the 8 other excluded induced subgraphs from Beineke's theorem.
To conclude, either no line graphs with integer Sombor index exist or they are likely very large.
