Problem 939

Solution:
Yes, there is such a set. One such set is $\{4,8,16,32,42\}$.
With this ticket, the subsums are all multiples of $4$ from $4$ to $60$, and all integers congruent to $2 \pmod{4}$ from $42$ to $102$. So we must show that any five numbers chosen by the lottery master will have a subsum equal to one of these.

We partition the numbers from $1$ to $50$ into three types: evens ($E$), large odds ($O$), and small odds ($o$). An odd is "large" if it exceeds $20$, otherwise "small". We argue by contradiction, starting from the assumption that the lottery master has found five numbers that will defeat the player.

First, some easy observations. Clearly any $E$'s chosen by the lottery master must be $2 \pmod{4}$, and must be no larger than $38$. The lottery master can't choose three $E$'s, since either two of them would add up to a multiple of $4$ from $4$ to $60$, or all three would add up to an integer that is $2 \pmod{4}$ from $42$ to $102$. It is also easy to see that any two $O$'s chosen by the lottery master must be in different classes $\pmod{4}$; in particular, the lottery master can't choose three $O$'s. Finally, any two $o$'s chosen by the lottery master must be in the same class $\pmod{4}$, and the lottery master can't choose four $o$'s.

The lottery master must choose at least one $E$. The only other possibility not yet ruled out is $OOOoo$, but in this case, the smallest $o$ combines with whichever $O$ is not in the same class modulo $4$ to make a multiple of $4$ less than or equal to $60$.

Now we have two overlapping cases: the lottery master can have $Eoo??$ or $EEoOO$.

- Case $Eoo??$: The $E$ and two $o$'s will add up to a multiple of $4$, which must be at least $64$. This implies that $E$ is $30, 34$, or $38$. We must avoid sums of the form $O+O$ that are $2 \pmod{4}$ and $\geq 42$, but we must also avoid sums of the form $E+o+O$ that are $2 \pmod{4}$ and $\leq 102$.
- If any $o$ is $19$, then these two considerations rule out all $O$ except $49$ (which requires that the $E$ is $38$).
- If any $o$ is $17$, then these considerations rule out all $O$ except $21$.
- If any $o$ is $15$, then these considerations rule out all $O$ except $23$.
- It isn't possible for all $o$ to be less than $15$.

Since we have at least two $o$'s, we can't have any $O$. We already ruled out four $o$'s, so we must have $EEooo$. But the smaller $E$ and the smallest two $o$'s will add up to a multiple of $4$ that's at most $60$, so this is a dead end.

- Case $EEoOO$: The two $E$'s must add up to at least $64$, so one of them is at least $34$. Pick the $O$ that's not congruent to the $o$ modulo $4$. Add these together, then add the smaller of the two $E$'s. The result is $2 \pmod{4}$, and is in the range $42$ to $102$, so we are finished.