Consider some distribution of $$$s$$$ problems such that the last element is $$$x>\lceil \frac s n \rceil$$$. Let's move some problems around to make the last element $$$x-1$$$. So there is a block of days with $$$x$$$ problems at the end. There also is a block of days with the number of problems equal to the first day. You can always move the problem from the first day of the ending block to the last day of the beginning block.

Let's call the first day of the ending block $$$r$$$ and the last day of the beginning block $$$l$$$. Observe that the condition between $$$r-1$$$ and $$$r$$$ never breaks. Since the difference between $$$a_{r-1}$$$ and $$$a_r$$$ decreased, if it was two-fold or less then it's still two-fold or less. Same for $$$l$$$ and $$$l+1$$$. In the case of $$$l=r-1$$$, the difference between $$$a_l$$$ and $$$a_r$$$ must have been at least $$$2$$$, thus, $$$a_l \le a_r$$$ after the move. The difference can't be $$$1$$$ because then it is exactly the same construction we have built for the minimum case, thus $$$x=\lceil \frac s n \rceil$$$. The condition between $$$r$$$ and $$$r+1$$$ also can't break. If $$$r$$$ is the last day, then there is no $$$r+1$$$. Otherwise, $$$a_r$$$ becomes equal to $$$a_{r+1} - 1$$$. Since all the values are positive, that difference can't become above two-fold. Same reasoning goes for $$$l$$$ and $$$l-1$$$.

So that move is always valid. That move also decreases the size of the ending block by $$$1$$$. Thus, at some point the ending block will be of size $$$1$$$, so making a move from it will decrease the last value by $$$1$$$. That proves that for any $$$x$$$ that can be the last value, $$$x-1$$$ can also be the last value. Thus, the function is indeed monotonous.

Consider the smallest value of $$$s$$$ that can be obtained: the last day has $$$x$$$ problems, the second to last has $$$\lceil \frac x 2 \rceil$$$ and so on.

In any construction there is a block of days with the number of problems equal to the first day. You can always add a problem to the last day of that block, thus increasing the total number of problems by $$$1$$$. The casework is basically the same as in the first proof.

That construction also implies that the smallest value of $$$s$$$ grows as $$$x$$$ grows.