A maharaja organized a children's festival. In the palace, a lot of gifts were hidden, which the children could search for and keep. The game ended when the children found all the gifts. To console the less fortunate children, each child received 22 rupees for each gift they did not find, minus 5 rupees for each gift they found. In the end, it turned out that every child found at least one gift, but no two children found the same number of gifts. The two least successful searching children together found 4 gifts, and the children received a total of 1957 rupees. How many gifts did the most successful child find?
Let the number of gifts be $a$, and the number of children be $g$. If the first child found $x_{1}$ items, then $a-x_{1}$ items were not found. Similarly, the second child found $x_{2}$ items and did not find $a-x_{2}$, and so on. The total number of items not found is $a-x_{1}+a-x_{2}+\ldots+a-x_{g}=g a-\left(x_{1}+x_{2}+\ldots+x_{g}\right)=g a-a$. Therefore, the Maharaja gave a total of $2(g a-a)$ rupees for the items not found, but withdrew $5 a$ rupees for the found items, and the difference between the two is 1957 according to the problem:

$$
2(g a-a)-5 a=1957
$$

from which

$$
a(2 g-7)=1957
$$

On the left side, both factors are positive and integers. Since the prime factorization of 1957 is 103 $\cdot$ 19, 1957 can only be factored into two positive integers in two ways: 103 $\cdot$ 19 or $1957 \cdot 1$. Since every child found at least one item, the number of gifts cannot be less than the number of children, so in the first case $\mathbf{a}=\mathbf{1 0 3}$ and $\mathbf{g}=\mathbf{1 3}$.

The two least successful children found four items, and there were no two children who found the same number of items, so one found 1 and the other 3 items. The remaining 11 children found the other 99. Since $4+5+6+\ldots+13+14=99$, and no other partition meets the requirements of the problem, the most successful searcher found 14 items in this case.

In the second case, $\mathbf{a}=\mathbf{1 9 5 7}$ and from $2 g-7=1$, $\mathbf{g}=\mathbf{4}$. The first two (least successful) children found 1 and 3 items respectively, and the other two found 1953 items together, so the most skilled one found at least 977, and since his partner also found at least four items, he found at most 1949 items. Every number between these two also meets the conditions. - Therefore, our problem has multiple solutions.

Note: No solver came up with the second possible partition.