Solution 1. As we are concerned with the relative position of Anthony and Clare, the answer is unchanged if Anthony remains stationary and Clare jumps $201$ lily pads to the right.
When does Clare next come close to Anthony? She makes one circuit of the pond after roughly $2020/201 \approx 10$ jumps, but in fact after those $10$ jumps she has travelled $2010$ pads so is $10$ lily pads short of Anthony. After $20$ jumps, she is $20$ lily pads short of Anthony, and so on. Each circuit her shortfall relative to Anthony goes up by $10$, until after $100$ jumps she is $100$ short of Anthony. But then after one more jump ($101$ in total) she is $101$ lily pads past Anthony.

And so it goes on; with each further loop Clare overshoots Anthony by $10$ pads less than the previous loop; after $111$ jumps, Clare overshoots Anthony by $91$; after $121$ Clare overshoots Anthony by $81$ and so on until after $201$ jumps, Clare overshoots Anthony by $1$ and they are on adjacent pads. Although the question asks for the first time Clare and Anthony are five or fewer pads apart, it turns out that the first time they are in this range, they are one pad apart.

Solution 2. After $t$ jumps, Clare is $201t$ pads clockwise from Anthony, measured modulo $2020$, so we have to solve:
$$201t \in \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \pmod{2020}.$$
We notice that $201^2 = 40401 \equiv 1 \pmod{2020}$ and so we can multiply both sides by $201$, to give:
$$t \in \{-1005, -804, -603, -402, -201, 0, 201, 402, 603, 804, 1005\} \pmod{2020}$$
The smallest $t > 0$ in this set is $t = 201$ and this solves the problem.