Lever

The key is to not get baited by the word "random". The random choices do not matter. Like, genuinely zero drama here.

For each position, define its excess as:

$$\max(0, a_i-b_i)$$

This is how many times that position still needs to be decreased before it is no longer greater than $b_i$.

Now look at the two steps.

Step 1: choose some index with $a_i>b_i$ and decrease it by $1$.

That always reduces the total excess by exactly $1$.

If $a_i-b_i=5$, it becomes $4$. If $a_i-b_i=1$, it becomes $0$. Either way, total positive excess drops by one.

Step 2: choose some index with $a_i<b_i$ and increase it by $1$.

This never increases the positive excess. Since the chosen value is below $b_i$, after adding $1$, it is still below or exactly equal to $b_i$. It cannot jump past $b_i$ and become greater. So step $2$ is basically irrelevant for deciding when step $1$ runs out of work.

So if initially

$$S=\sum_{i=1}^{n}\max(0,a_i-b_i),$$

then step $1$ succeeds exactly $S$ times. After those $S$ successful decreases, there is no index left with $a_i>b_i$.

But the Lever does not stop immediately after the last successful decrease. It starts one more iteration. In that final iteration, step $1$ is ignored, and only after the iteration does the Lever stop.

Therefore the answer is:

$$S+1$$

That extra $+1$ is the whole trick. Miss it and the sample punches you in the face.