What is the value of this "game" (counters rebalancing)?
Here is an upper bound for question 2.   

Theorem: After $TM$ steps, you can have at most $T \log M$ positive counters. 

Notation: we assign each step a number, ending with the $0$th step, and starting with the $(-TM+1)$th step.

Observation: in the optimal strategy, if you increment a counter, you do not let it return to $0$.

Proof: Let us assign a value of $1$ to each time you increment a counter in the last $T$ steps; $\frac{1}{2}$ to each time you increment a counter in the previous $T$ steps; $\frac{1}{3}$ to each time you increment a counter in the $T$ steps before that, etc. That is, if you increment a counter between step $-kT+1$ and $-(k-1)T$, it gets a value of $\frac{1}{k}$. 
We need a lemma that we will prove later.

Lemma: for a counter to be positive, it must have had a total value of at least $1$ assigned to it. 

With the lemma, the proof is easy:  the total value we have to assign is $T \sum_{i=1}^M \frac{1}{i} \approx T \log M$, and each positive counter requires us to have assigned total value of at least $1$ to it.

Proof of Lemma: Look at the first time you made that counter positive. If that was between $-kT$ and $-(k-1)T$, then it has been decremented at least $k-1$ times, and so you need to have incremented it $k$ times to have it end up positive. But each of these increments assigns it a value of at least $\frac{1}{k}$.

And here is an upper bound for question 1. Let's use the same overall proof strategy as before.  Again, we assign a value of $\frac{1}{k}$ when we increment a counter on a step between $-kT+1$ and $-(k-1)T$. We also assign a value of $\frac{1}{M}$ when we increment a counter on a step between $-2MT$ and $-MT$, and a value of $0$ if we increment a counter before that. By the same reasoning as before, any counter that was $0$ at step $-2MT$ must be assigned a value of at least $1$. But now, if a counter was positive at step $-2MT$, it still must have been assigned a total value of at least $1$, since it must have been incremented $M$ times to end up positive. Now, the total value we have to assign is $T\log M + T$, so we can have at most $T (\log M + 1)$ positive counters.