### Part 1:
1. Given $a_0 = 24$, we need to find the smallest $n$ such that $a_n = 0$.
2. We calculate the sequence step-by-step:
$$\begin{aligned} &a_0 = 24, \\ &a_1 = a_0 - \lfloor \sqrt{a_0} \rfloor = 24 - \lfloor \sqrt{24} \rfloor = 24 - 4 = 20, \\ &a_2 = a_1 - \lfloor \sqrt{a_1} \rfloor = 20 - \lfloor \sqrt{20} \rfloor = 20 - 4 = 16, \\ &a_3 = a_2 - \lfloor \sqrt{a_2} \rfloor = 16 - \lfloor \sqrt{16} \rfloor = 16 - 4 = 12, \\ &a_4 = a_3 - \lfloor \sqrt{a_3} \rfloor = 12 - \lfloor \sqrt{12} \rfloor = 12 - 3 = 9, \\ &a_5 = a_4 - \lfloor \sqrt{a_4} \rfloor = 9 - \lfloor \sqrt{9} \rfloor = 9 - 3 = 6, \\ &a_6 = a_5 - \lfloor \sqrt{a_5} \rfloor = 6 - \lfloor \sqrt{6} \rfloor = 6 - 2 = 4, \\ &a_7 = a_6 - \lfloor \sqrt{a_6} \rfloor = 4 - \lfloor \sqrt{4} \rfloor = 4 - 2 = 2, \\ &a_8 = a_7 - \lfloor \sqrt{a_7} \rfloor = 2 - \lfloor \sqrt{2} \rfloor = 2 - 1 = 1, \\ &a_9 = a_8 - \lfloor \sqrt{a_8} \rfloor = 1 - \lfloor \sqrt{1} \rfloor = 1 - 1 = 0. \end{aligned}$$
3. Therefore, the smallest $n$ such that $a_n = 0$ is $\boxed{9}$.

### Part 2:
1. We claim that $a_{2j} = (m-j)^2$ for all $1 \leq j \leq m$. We will prove this by induction.
2. Base Case:
$$\begin{aligned} &a_0 = m^2, \\ &a_1 = a_0 - \lfloor \sqrt{a_0} \rfloor = m^2 - m, \\ &a_2 = a_1 - \lfloor \sqrt{a_1} \rfloor = (m^2 - m) - (m - 1) = (m - 1)^2. \end{aligned}$$
Thus, the base case holds.
3. Inductive Step:
Assume $a_{2k} = (m-k)^2$ for some $k$. We need to show $a_{2k+2} = (m-k-1)^2$.
$$\begin{aligned} &a_{2k+1} = a_{2k} - \lfloor \sqrt{a_{2k}} \rfloor = (m-k)^2 - (m-k) = m^2 - m(2k+1) + k^2 + k, \\ &a_{2k+2} = a_{2k+1} - \lfloor \sqrt{a_{2k+1}} \rfloor. \end{aligned}$$
Since $(m-k-1)^2 \leq a_{2k+1} \leq (m-k)^2$, we have $\lfloor \sqrt{a_{2k+1}} \rfloor = m-k-1$.
$$\begin{aligned} &a_{2k+2} = a_{2k+1} - (m-k-1) = (m^2 - m(2k+1) + k^2 + k) - (m-k-1) \\ &= m^2 - m(2k+2) + k^2 + 2k + 1 = (m-k-1)^2. \end{aligned}$$
Thus, the induction is complete.
4. Using the fact that $a_{2j-2} = (m-j+1)^2$ for all $2 \leq j \leq m$, we have:
$$\begin{aligned} &a_{2j-1} = a_{2j-2} - \lfloor \sqrt{a_{2j-2}} \rfloor = (m-j+1)^2 - (m-j+1) \\ &= (m-j+1)(m-j). \end{aligned}$$
Therefore, $\boxed{a_{2j-1} = (m-j+1)(m-j)}$ and $\boxed{a_{2j} = (m-j)^2}$.

### Part 3:
1. Let $a_0 = m^2 - p$. We need to find $k$ such that $a_k = (m-p)^2$.
2. We calculate the sequence step-by-step:
$$\begin{aligned} &a_0 = m^2 - p, \\ &a_1 = a_0 - \lfloor \sqrt{a_0} \rfloor = m^2 - p - \lfloor \sqrt{m^2 - p} \rfloor. \end{aligned}$$
Since $\lfloor \sqrt{m^2 - p} \rfloor = m - 1$ (as $p$ is small), we have:
$$\begin{aligned} &a_1 = m^2 - p - (m - 1) = m^2 - m + 1 - p. \end{aligned}$$
3. We continue this process until we reach $(m-p)^2$. The sequence will eventually reach $(m-p)^2$ after $2p$ steps.
4. Therefore, the smallest $n$ such that $a_n = 0$ is $\boxed{2m - 1}$.