Mathematical Olympiad Rioplatense

Having $108$ pesos is sufficient. There is only one way to express $155$ as $11a+13b$ with $a$, $b$ nonnegative integers: with $a=7$ and $b=6$. So each customer has $7$ coins of $11$ and $6$ coins of $13$. Hence to serve a customer as desired it is enough to have available either $6$ coins of $11$ or $5$ coins of $13$ (or both). In the first case the salesperson takes from the customer $6$ coins of $13$ and gives back $6$ coins of $11$: $6\cdot 13-6\cdot 11=12$. In the second case the salesperson takes $7$ coins of $11$ and gives back $5$ coins of $13$: $7\cdot 11-5\cdot 13=12$.

An initial amount of $N\ge 108$ pesos ensures $6$ coins of $11$ or $5$ coins of $13$. Indeed if there were at most $5$ coins of $11$ and at most $4$ coins of $13$ then $N\le 5\cdot 11+4\cdot 13=107$. So the first customer can be attended whenever $N\ge 108$. The remaining ones can be attended too, because $N$ will be still greater when their turn comes.