The full solution I have to this problem is full of details and edge cases, so I skim over them. If you want to follow along a picture is strongly recommended. Let $m=\lceil n/k \rceil$. Put the grid in the coordinate plane with bottom left corner at the origin, and draw the lines $x=k,x=2k,\ldots,x=(m-1)k$ and $y=k,y=2k,\ldots,y=(m-1)k$, splitting the grid into $m^2$ rectangles. Assign a charge to every rectangle: $1$ for every snake completely inside, $0.5$ for every snake partially inside. First, I claim that every rectangle has charge at least $0.5$, i.e. no rectangle is empty, with the possible exception of the "small" square that's $(n-(m-1)k) \times (n-(m-1)k)$ in the corner, unless $k \mid n$ (in which case the square is just as large as everything else). The idea is to prove that if some rectangle has some snakes in it (possibly partially) then any rectangle sharing a border of length $k$ with it also has snakes in it. WLOG, suppose otherwise let the first rectangle be below the second rectangle. Then consider the topmost python in the first rectangle, which must exist since if we have an anaconda there then there's a python either in the first or second rectangle. If the topmost python isn't fully contained in the first rectangle then we either find that some anaconda stretches into the second rectangle, or that there exists some python above it, either inside the second rectangle or the first, which are both contradictions. On the other hand, if the topmost python is fully contained, then take some horizontally adjacent rectangle to the first, WLOG on the right, which must contain an anaconda in its leftmost column. This anaconda must either have an adjacent python extending into the second rectangle at the end, have a python extending into the first rectangle above the topmost python, or have a python fully contained in the same rectangle. The first two cases are obvious contradictions, and in the third case the python has an anaconda on the rightmost column of the first rectangle that extends into the second. Now the idea is to show that all these rectangles have charge at least $1$. For the $k \times k$ rectangles this is obvious, since if there's a snake partially inside then there must be another. The issue is that this will break for our $(n-(m-1)k) \times k$ rectangles along the edge (not when $k \mid n$), since we could run into a wall. This can be dealt with by essentially moving the charge over to the $k \times k$ rectangle adjacent to it: if an $(n-(m-1)k) \times k$ rectangle, say on the top edge of the grid, has charge only $0.5$, then the adjacent rectangle must have some python. If this python isn't fully contained inside the $k \times k$ then there must be an anaconda partially (or fully) in it too (not the one also in the $(n-(m-1)k) \times k$), so the charge of the $k \times k$ rectangle is at least $1.5$. Otherwise, if the python is fully contained, the conclusion is immediate. There's one more edge case, which occurs when we look at the two $(n-(m-1)k) \times k$ rectangles adjacent to the $(n-(m-1)k) \times (n-(m-1)k)$ corner piece. If at most one of these has charge $0.5$ we are done by the same argument, but if both have charge $0.5$ then we want to prove that the $k \times k$ square bordering them both must have charge at least $2$. But this is immediate if the two snakes in the $(n-(m-1)k) \times k$ rectangles don't overlap, because then each must have another snake touching it for a total of $4$ snakes at least partially inside the $k \times k$ rectangle. But if they do overlap, we still need at least one snake adjacent, and if it's fully inside, then we're done, otherwise that adjacent snake produces at least one more snake adjacent to the "adjacent snake" itself (or lies in an $(n-(m-1)k) \times k$ rectangle, which contradicts our assumption). Now we have to show that if $n \nmid k$ then we can distribute the charge of $1$ that should be in the $(n-(m-1)k) \times (n-(m-1)k)$ corner square somewhere. Most of the time we can actually show that the top right four rectangles will hold charge at least $4$ in a very similar way to what we did for our "last edge case", so we're done. However, there is a case where: The top rightmost small square is empty The two rectangles adjacent to it have charge $0.5$, and their snakes share some positive perimeter in the $k \times k$ square The snake (WLOG a python) that must be adjacent to the aforementioned snakes is fully contained in the $k \times k$ square This snake requires an anaconda adjacent to it, which is partially contained in one of the two $(n-(m-1)k) \times k$ rectangles but not completely In this case we only get a charge of $3.5$. But note that the snake mentioned in the third bullet point requires an anaconda along the rightmost column of the $k \times k$ square to the left of the four rectangles we are considering (which exists because otherwise we wouldn't be able to fit the anaconda in the fourth bullet point). If this anaconda is partially contained in the $(n-(m-1)k) \times k$ rectangle above the left $k \times k$ square then we find that the six rectangles we mentioned have charge at least $6.5$, since this anaconda cannot touch the top of the grid. Otherwise, the four original rectangles and the left $k \times k$ square together have a charge of at least $5$, since the anaconda contributes at least $0.5$ and there's also a python on top of it that's fully contained in the two $k \times k$ squares we consider. Putting everything together, we find that the total charge is at least $m^2$. But the total charge is exactly the number of snakes there are, so there are at least $m^2$ snakes and the sum of their squared lengths is thus at least $m^2k^2 \geq n^2$, as desired. $\blacksquare$

Call any valid configuration \emph{poisonous}. Call a snake which touches an outer edge of the grid but is not parallel to that edge \emph{hurt}. We will prove by induction on $n$ that in any poisonous configuration the sum of the lengths of the hurt snakes touching each side is at least $2n$. Consider a $n+2 \times n+2$ grid $G$, and the $n \times n$ grid $H$ formed by removing the outer edges of $G$. We now apply the following operation to snakes parallel to an outside edge of the grid and on the outside edge of the grid (call such snakes \emph{lazy}): if they are not in a corner or in a square adjacent to a corner, move them in by one space (possibly merging them with any snakes which they would overlap with). If they are one space away from the corner, move them in by one space but delete the part of that snake which moved to a corner of $H$. If they are in a corner move them up one space and delete the part of that snake moving from a corner of $G$. The resulting configuration in $H$ is poisonous. Hence, the sum of hurt snakes colliding with each edge is at least $2n$. However, note that any hurt snake has to be hurt in $G$ as well. However, note that every valid configuration either has a snake taking up an entire row or column or hurt snakes for every outside edge. In the first case we are done as we would then get a snake taking up a whole row. In the second case we are also done, as the sum is at least $2n+4 = 2(n+2)$. Now, another induction finishes.