We will show that he can do so by the morning of day $199$ but not earlier.

If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$V = \begin{cases} n & m = 0, \\ n + \frac{1}{2} & m = 1, \\ n + 1 & m \ge 2. \end{cases}$$
We also denote this position by $(n, m)$. Implicitly we will also write $k$ for the number of piles with exactly two pieces of rubbish.

Angel's strategy is the following:
(i) From position $(0, m)$ remove one piece from each pile to go position $(0, 0)$. The game ends.
(ii) From position $(n, 0)$, where $n \ge 1$, remove one pile to go to position $(n - 1, 0)$. Either the game ends, or the demon can move to position $(n - 1, 0)$ or $(n - 1, 1)$. In any case $V$ reduces by at least $1/2$.
(iii) From position $(n, 1)$, where $n \ge 1$, remove one pile with at least two pieces to go to position $(n - 1, 1)$. The demon can move to position $(n, 0)$ or $(n - 1, 2)$. In any case $V$ reduces by (at least) $1/2$.
(iv) From position $(n, m)$, where $n \ge 1$ and $m \ge 2$, remove one piece from each pile to go to position $(n - k, k)$. The demon can move to position $(n, 0)$ or $(n - k, k + 1)$. In any case $V$ reduces by at least $1/2$. (The value of position $(n - k, k + 1)$ is $n + \frac{1}{2}$ if $k = 0$, and $n - k + 1 \le n$ if $k \ge 1$.)

So during every day if the game does not end then $V$ is decreased by at least $1/2$. So after $198$ days if the game did not already end we will have $V \le 1$ and we will be in one of positions $(0, m)$, $(1, 0)$. The game can then end on the morning of day $199$.

We will now provide a strategy for demon which guarantees that at the end of each day $V$ has decreased by at most $1/2$ and furthermore at the end of the day $m \le 1$.
(i) If Angel moves from $(n, 0)$ to $(n - 1, 0)$ (by removing a pile) then create a new pile with one piece to move to $(n - 1, 1)$. Then $V$ decreases by $1/2$ and $m = 1 \le 1$.
(ii) If Angel moves from $(n, 0)$ to $(n - k, k)$ (by removing one piece from each pile) then add one piece back to each pile to move to $(n, 0)$. Then $V$ stays the same and $m = 0 \le 1$.
(iii) If Angel moves from $(n, 1)$ to $(n - 1, 1)$ or $(n, 0)$ (by removing a pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1/2$ and $m = 0 \le 1$.
(iv) If Angel moves from $(n, 1)$ to $(n - k, k)$ (by removing a piece from each pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1/2$ and $m = 0 \le 1$.

Since after every move of demon we have $m \le 1$, in order for Angel to finish the game in the next morning we must have $n = 1, m = 0$ or $n = 0, m = 1$ and therefore we must have $V \le 1$.

But now inductively the demon can guarantee that by the end of day $N$, where $N \le 198$ the game has not yet finished and that $V \ge 100 - N/2$.

Define Angel's score $S_A$ to be $S_A = 2n + m - 1$. The Angel can clear the rubbish in at most $\max\{S_A, 1\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.

Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_A$ reduces by at least $1$ in each day. (Case (i) is trivial as the game ends in one day.)

Now define demon's score $S_D$ to be $S_D = 2n - 1$ if $m = 0$ and $S_D = 2n$ if $m \ge 1$. The claim is that if $(n, m) \ne (0, 0)$, then the demon can ensure that Angel requires $S_D$ days to clear the rubbish.

Again, demon's strategy is the same as in the Solution by PSC and in each of cases (i)-(iv) one needs to check that $S_D$ reduced by at most $1$ in each day.