Let there are not such problem. Then every problem is solved by all members( type A) or by $\leq 19$ members(type B). There are $40*4=160$ wrong solutions, so there are $\leq \frac{160}{21}<8$ problems with type B. But also obvious, that there are $\geq 5$ problems with type B. And so there are $30-|B|$ problems with type $A$. So every member solved $26-(30-|B|)=|B|-4$ problems with type $B$. But $\binom{n}{n-4}=1,15,35<40$(for $n=5,6,7$) so some members solved same set of problems - contradiction.

Suppose there are not such problem. Simple double counting 1. There $26*40=1040$ solved problems. 2. Suppose $X$ problems - solved by all members and $30-X$ by at most $19$. So there are less than $X*40+(30-X)*20=20*X+600$ solved problems. So from (1), $20X > 440$ so $X>22$. Contradiction because there will be two members with same problem sets.

Assume the contrary; then every problem was solved by either at most $19$ jury members or all $40$. Suppose there are $n$ problems solved by all $40$, where $n$ is a nonnegative integer. There are $30-n$ problems left to be solved by $19$ or less jury members. Each jury member must solve $26-n$ more times. So there are $40(26-n)$ solves left to be done by the jury, and a maximum of $19(30-n)$ solves that can be done on the $30-n$ problems left. This gives $$40(26-n) < 19(30-n),$$so the only positive integer values of $n$ that don't break the rule that every member must solve $26$ problems is $n>22$. Each jury member must also solve a distinct set of problems. This means that of the $30-n$ problems that weren't solved by all $40$ jury members, there must be at least $40$ distinct sets of $26-n$ of the problems. So $$\binom{30-n}{26-n} > 40.$$The solutions to this inequality over the integers are $n<23$ and $n>34$. We have already established that if $n$ is a positive integer, it is greater than $22$, so the solution $n<23$ is invalid in these circumstances. But there are only $30$ problems so clearly it is impossible for more than $30$ problems to be solved. Hence the only value of $n$ that satisfies our conditions is $n=0$; there can be no problems that were solved by all $40$ members of the jury. So if our initial assumption is true, each of the $30$ problems were solved at most $19$ times. This gives that the maximum number of solves is $30\cdot 19=570$. But since each jury member solves $26$ problems, there should be $26\cdot40=1040$ solves. This is another contradiction, so we conclude a) it is impossible for any of the $30$ problems to be solved by all forty members of the jury and b) it is impossible for all of the $30$ problems to be solved by less than half of the jury. Thus we conclude that there must exist a problem that was solved by at least half the jury but not all of it, and we are done.