First, lets give a short story about the problem which will follow:
I was at school here in Greece. We have school 5 days of the week. 2 random days we have 7 hours of lessons and the rest 6 hours of lessons. I wanted to return home, when possible, with one of my friends. However, we are not in the same class, so he has a different schedule. By taking all cases, I noticed that there will be at least 1 day of the week in which we will have the same hours of lessons, so we can return together.
I then thought of a more general statement:
If we arrange $n$ in number $A$ letters and $n - 1$ in number $B$ letters with $2$ ways, then there exists at least one same letter in the same place, if we assume that the first term for every arrangement has place 1.
I though about it myself and could not figure out a rigorous proof. Lastly, I posted it on a discord server dedicated to Olympiad Math and got this response:
"well... after we're done with the first way to arrange it a second way we could essentially swap the B's with (n-1) A's, thus leaving 1 A at the same place so I would guess this is true."
Although I understand what is being said, I do not understand how this is enough to prove the original statement. So, if you happen to have a rigorous proof or some clever ideas, please share!
On your original example, between you there are four $7$-hour days, so they can occur at most on four different days of the $5$-day week, so at least $5-4=1$ day must have both of you having $6$-hour days.
In your generalised form in the two arrangements of $2n-1$ letters ($n$ copies of A and $n-1$ of B) you can say the same thing: between the two arrangements, the Bs can appear in at most $2(n-1)$ different positions, so at least $(2n-1)-2(n-1)=1$ position must have As in both arrangements.
A pigeonhole argument would say there are $2n$ As across the two arrangements to put in $2n-1$ positions so at least one position has 2 As. This is essentially the same argument.