I think you simply do not understand my argument. Let's use smaller numbers. Suppose Peter does at least one problem a day for 7 days and at most 10 problems in those seven days. You claim that because

and

, there might not exist an interval of days such that Peter does exactly 4. I can prove to you that there is such an interval of days. There are four cases to consider.

Case 1:

. Then there are 3 number

that are greater than 4. But then,

,

,

. That means there are six values, all in the range

. Now,

only if

. So, we consider only two cases:

and

. If

, then

are six values in the range

, so there must be two that are equal. If

, then

are seven values in the range

. Either way, at least two must be equal.

Case 2:

. Then there are four numbers:

that are greater than 4. But then,

all must be less than or equal to

since

, so

and

, so

. Again, you have seven values in a range of only 6 values.

Case 3:

. Then there are five numbers:

that are greater than 4. But then

must both be less than or equal to

. Again, there are seven numbers in a range of only six values, so at least two must be equal.

Case 4:

. Then there are six numbers:

that are greater than 4. But, still,

, so there are still at least seven values in a range of only six values, so at least two must be equal.

Those are the only possible cases. This same proof can be extended to find an interval of days where Peter solves exactly 5 or 6 problems, and as I said, I believe it can even be extended to find an interval of days where Peter solves exactly 7 problems.