Answer: 3750 rubles.

Solution. Let's call a 4-week period a "month", there are 13 such "months" in a year. If all the bills given by Masha were hundred-ruble bills, her sister would have received 1300 rubles. This means Masha gave 100 rubles twelve times and 50 rubles once. If in any "month" Masha gave 100 rubles, it means that in her piggy bank there were only hundreds. So, over these 12 "months" she kept 12 * 300 = 3600 rubles for herself.

Thus, 50-ruble bills could appear in Masha's possession only in one month out of 13. If in a "month" only one 50-ruble bill was put into the piggy bank, she gave it to her sister at the end of the month, so her "income" remained 300 rubles.

If in any "month" Masha saves not less than two 50-ruble bills, she gives them to her sister over consecutive "months", which contradicts the condition. The exception is the case when they all come in the 13th "month", then she won't have time to give them away. So, in this case, for the first 12 "months" Masha received 300 rubles each, and in the last one, she could put in the piggy bank from zero to three 50-ruble bills, that is, she could fall short of 300 rubles by a maximum of 150 rubles.

Grading criteria: It is stated that Masha gave money 13 times, 12 times 100 rubles and once 50 rubles - 5 points. The correct answer without mentioning that the 50 rubles are given only in the last month - 10 points. The correct solution to the problem, but the answer indicates the sum including the money given to her sister (3750 + 1250 = 5000) - 17 points.
Task 1. Masha has a piggy bank where she puts a 50 or 100 ruble bill every week. At the end of every 4 weeks, she picks the bill of the smallest denomination from the piggy bank and gives it to her little sister. After a year, it turned out that she had given her sister 1250 rubles. What is the minimum amount of money that could have accumulated with her over this time?