A really nice problem. Answer. $\boxed{\text{Yes.}}$ Name the first knight, Lancelot, the second knight, Gawain, and the third one as Percival. Lancelot's attack reduces the number of heads from $x$ to $x-\frac{x}{2}-1=\frac{x}{2}-1$. Gawain's attack reduces the number of heads from $x$ to $x-\frac{x}{3}-2=\frac{2x}{3}-2=2\left(\frac{x}{3}-1\right)$. Percival's attack reduces the number of heads from $x$ to $x-\frac{x}{4}-3=\frac{3x}{4}-3=3\left(\frac{x}{4}-1\right)$. Now, we describe an algorithm that reduces the number of heads and therefore eventually we must hit $0$. Denote a current number of heads as x. We describe some moves: Percival + Gawain Duo. If $x=4k$, then we reduce $x$ to $2(k-2)$ using Percival and then Gawain. Percival's Slam. If $x=4k$, then we reduce $x$ to $3(k-1)$ using only Percival. Lancelot's Special Attack. If $x=2r$, then we reduce $x$ to $r-1$ using only Lancelot. Note that if $r=k-2$, then we reduce $x$ to $k-3$. As at the beginning $4\mid x=41!$, we use Percival + Gawain Duo. Note that if $k$ is even after that move, we repeat using Percival + Gawain Duo until $k$ is odd. If $k\equiv 1\pmod{4}$ after some point, then we use Percival's Slam instead and we obtain new $x$ divisible by $4$ and repeat with Percival + Gawain Duo again. If $k\equiv 3\pmod{4}$, then we use Lancelot's Special Attack after Percival + Gawain Duo and we obtain new $x=k-3\equiv 0\pmod{4}$ and we repeat with Percival + Gawain Duo again. We are done since at some point, we must eventually hit $0$.