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**SlipEternal** 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 $\displaystyle 7+7 = 14$ and $\displaystyle 10+3 = 13$, 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: $\displaystyle S_4 = 4$. Then there are 3 number $\displaystyle S_5,S_6,S_7$ that are greater than 4. But then, $\displaystyle S_1 + 4 = 5 < 7\le S_7$, $\displaystyle S_2 + 4 = 6 < 7 \le S_7$, $\displaystyle S_3 + 4 =7 \le S_7$. That means there are six values, all in the range $\displaystyle \{5, \ldots, S_7\}$. Now, $\displaystyle S_4+4 \le S_7$ only if $\displaystyle S_7 \ge 8$. So, we consider only two cases: $\displaystyle S_7 = 7$ and $\displaystyle S_7 \ge 8$. If $\displaystyle S_7 = 7$, then $\displaystyle \{S_5,S_6,S_7,S_1+4,S_2+4,S_3+4\}$ are six values in the range $\displaystyle \{5,6,7\}$, so there must be two that are equal. If $\displaystyle S_7 \ge 8$, then $\displaystyle \{S_5,S_6,S_7,S_1+4,S_2+4,S_3+4,S_4+4\}$ are seven values in the range $\displaystyle \{5,6,7,8,9,10\}$. Either way, at least two must be equal.

Case 2: $\displaystyle S_3\le 4, S_4>4$. Then there are four numbers: $\displaystyle S_4,S_5,S_6,S_7$ that are greater than 4. But then, $\displaystyle S_1+4,S_2+4,S_3+4$ all must be less than or equal to $\displaystyle S_7$ since $\displaystyle S_7 \ge S_4+3 >7$, so $\displaystyle S_7 \ge 8$ and $\displaystyle S_3 \in \{3,4\}$, so $\displaystyle S_3+4 \le 8$. Again, you have seven values in a range of only 6 values.

Case 3: $\displaystyle S_2 \le 4, S_3>4$. Then there are five numbers: $\displaystyle S_3,\ldots,S_7$ that are greater than 4. But then $\displaystyle S_1+4,S_2+4$ must both be less than or equal to $\displaystyle S_7$. Again, there are seven numbers in a range of only six values, so at least two must be equal.

Case 4: $\displaystyle S_1 \le 4, S_2>4$. Then there are six numbers: $\displaystyle S_2,\ldots, S_7$ that are greater than 4. But, still, $\displaystyle S_1+4 \le S_7$, 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.