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.