1. Assume the contrary: Suppose that every problem was solved by either at most 19 jury members or by all 40 jury members. Let $n$ be the number of problems solved by all 40 members, where $n$ is a nonnegative integer.

2. Remaining problems: There are $30 - n$ problems left to be solved by at most 19 jury members. Each jury member must solve $26 - n$ problems from these remaining problems.

3. Counting total solves: The total number of solves by the jury members is $40(26 - n)$. The maximum number of solves for the $30 - n$ problems is $19(30 - n)$. Therefore, we have the inequality:
$$40(26 - n) \leq 19(30 - n)$$
Simplifying this inequality:
$$1040 - 40n \leq 570 - 19n$$
$$1040 - 570 \leq 40n - 19n$$
$$470 \leq 21n$$
$$n \geq \frac{470}{21} \approx 22.38$$
Since $n$ must be an integer, $n \geq 23$.

4. Distinct sets of problems: Each jury member must solve a distinct set of problems. For the $30 - n$ problems that weren't solved by all 40 jury members, there must be at least 40 distinct sets of $26 - n$ problems. This gives the inequality:
$$\binom{30 - n}{26 - n} > 40$$
Simplifying the binomial coefficient:
$$\binom{30 - n}{4} > 40$$
The solutions to this inequality over the integers are $n < 23$ and $n > 34$. However, we have already established that $n \geq 23$, so the solution $n < 23$ is invalid. Also, since there are only 30 problems, $n > 34$ is impossible.

5. Contradiction: The only value of $n$ that satisfies our conditions is $n = 0$. This means there can be no problems that were solved by all 40 members of the jury.

6. Maximum number of solves: If our initial assumption is true, each of the 30 problems were solved at most 19 times. This gives a maximum number of solves:
$$30 \cdot 19 = 570$$
But since each jury member solves 26 problems, the total number of solves should be:
$$26 \cdot 40 = 1040$$
This is another contradiction.

7. Conclusion: We conclude that it is impossible for any of the 30 problems to be solved by all 40 members of the jury and it is impossible for all of the 30 problems to be solved by less than half of the jury. Thus, there must exist a problem that was solved by at least half the jury but not all of it.

$\blacksquare$