Answer: The desired minimum is $N$ if $N$ is odd, and $(N-1)$ if $N$ is even.
We say a drone covers a vertex if it takes a $90^\circ$ turn at that vertex.

Construction: Let $v_1, \dots, v_N$ denote the $N$ vertical lines and $h_1, \dots, h_N$ denote the $N$ horizontal lines of the grid.
For $N = 2n$, make one drone go along the outer boundary of the grid, and let $(n-1)$ of the drones traverse the rectangles enclosed by $v_{2i}, v_{2i+1}, h_1, h_N$ for all $1 \le i \le n-1$, and let $(n-1)$ of the drones traverse the rectangles enclosed by $h_{2i}, h_{2i+1}, v_1, v_N$ for all $1 \le i \le n-1$. These $N-1$ drones cover all roads in Sibyl.
For $N = 2n + 1$, again make one drone go across the boundary, and let $n$ of the drones traverse the rectangles enclosed by $v_{2i}, v_{2i+1}, h_1, h_N$ for all $1 \le i \le n$, and let $n$ of the drones traverse the rectangles enclosed by $h_{2i}, h_{2i+1}, v_1, v_N$ for all $1 \le i \le n$. One notes that these $N$ drones cover all roads in Sibyl.

Estimate: Suppose $k$ drones suffice. We will first show for all $N$ that $k \ge N-1$. Fix the top-left corner $A$ of the grid and call any drone that passes through this town to be cornered. Suppose $C$ is the set of drones that are cornered. Let $\mathcal{L}_h, \mathcal{L}_v$ denote the set of drones (not in $C$) that cross a street in $h_1, v_1$ respectively. Note that any drone that covers some vertex in $h_1$ is either in $C$ or $\mathcal{L}_h$ and any drone in $\mathcal{L}_h$ covers two vertices in $h_1$ and any drone in $C$ covers one vertex in $h_1$ other than $A$. Since each of the $N$ vertices on the top edge needs to be covered, $1 + |C| + 2|\mathcal{L}_h| \ge N$. Similarly, $1 + |C| + 2|\mathcal{L}_v| \ge N$. Adding, we get $2 + 2|C| + 2|\mathcal{L}_h| + 2|\mathcal{L}_v| \ge 2N$, hence $k \ge |C| + |\mathcal{L}_h| + |\mathcal{L}_v| \ge N-1$, proving the claim.

Now suppose $N$ is odd. If possible, suppose $k = N-1$. Following the notation in the last part, we see that equality must hold: $1 + |C| + 2|\mathcal{L}_h| = N$ and $1 + |C| + 2|\mathcal{L}_v| = N$. Further, every drone must be in one of these three sets, and no point on the top edge apart from the top-left corner can be covered by two drones (else the bounds would not be tight).

Call a drone a dominator if it passes through both $h_1, h_N$ (a vertical dominator) or both $v_1, v_N$ (a horizontal dominator). Since every drone passes through either $h_1$ or $v_1$, and this reasoning applies any other corner, this implies that every drone must be a dominator. Indeed, if some drone is not a dominator, one can pick one of $h_1, h_N$ and one of $v_1, v_N$ so that it doesn't pass through any of the picked lines; and applying the above reasoning to the corner at the intersection of these two leads to a contradiction.

Now if there are two horizontal dominators through $A$, they both cover the top-right corner, contradiction. Similarly, there can be at most one vertical dominator through $A$. But $|C|$ is at least one (some drone needs to cover $A$), and $|C| = 1$ leads to a contradiction modulo $2$ in $1 + |C| + 2|\mathcal{L}_h| = N$, so $|C|$ is exactly $2$, and there is a horizontal and a vertical dominator through $A$.

By a similar reasoning, there is a vertical and a horizontal dominator through the bottom-right corner, $D$. But the horizontal dominator through $A$ and the vertical dominator through $D$ both cover the top-right corner point. We noted before that no point on the top edge apart from the top-left corner can be covered by two drones, so this is a contradiction, showing $k \ge N$. $\square$