2 In the parliament, there are 2000 deputies who decide to review the budget, which consists of 200 expenditures. Each deputy prepares a budget draft, listing the amounts for each expenditure. The total amount listed by each does not exceed S. During the parliamentary review, each amount is set to a number agreed upon by at least k deputies (i.e., the amount listed by at least k deputies is not less than the approved amount). What is the minimum value of k that ensures the total approved amount does not exceed S? (54th Moscow Mathematical Olympiad problem)
2. Obviously, the expenses filled by each member of parliament form a sequence of length 200, and the 2000 sequences filled by the members of parliament constitute a $2000 \times 200$ number table. The sum of each row in this table does not exceed $S$. Now, a row of numbers needs to be filled at the bottom of the table, such that each number in this row is less than at least $k$ numbers in the corresponding column. Find the smallest $k$ such that the sum of the numbers in the last filled row does not exceed $S$. If a number $k$ meets the above requirements, it is called good. $k=2000$ is clearly good. Furthermore, $k=1999$ is also good. In fact, when $k=1999$, there is only one number in each column that is less than $x_{j}$, and we call such a number a bad number. Thus, there are at most 200 bad numbers in 200 columns. However, the table has 2000 rows, so at least one row has no bad numbers. This indicates that the member of parliament who filled this row agrees with every expense selected. Continuing this process (constantly narrowing the scope), it is not difficult to find that $k=1991$ is good. In fact, when $k=1991$, there are at most 9 bad numbers in each column, and 200 columns have at most 1800 bad numbers. Therefore, at least one row has no bad numbers. Below, we prove that $k=1990$ is not good. In fact, divide the table into groups of 10 rows each. In the $i$-th group, each row is filled with $\frac{S}{199}$, $\frac{S}{199}, \cdots, \frac{S}{199}, 0, \frac{S}{199}, \cdots, \frac{S}{199}$. Only the $i$-th column is 0, and the rest are $\frac{S}{199}$. Thus, each row in the table has 199 $\frac{S}{199}$ and one 0. Therefore, the sum of each row is $S$. Each column has 10 zeros and 1990 $\frac{S}{199}$. Thus, for any $j$, let $x_{j}=\frac{S}{199}$. The total approved expenditure is $200 \cdot \frac{S}{199}>S$, which is a contradiction. In summary, the minimum value of $k$ is 1991.