Answer: $14$

Note that $74$ and $111$ have a common factor of $37$. Also, $87$ and $124$ are $13$ greater than $74$ and $111$ respectively. We call the coins with denominations $87$ and $124$ 'bad coins'.

Since $2023 \equiv 25 \pmod{37}$, we must use $k$ bad coins such that $13k \equiv 25 \pmod{37}$. The smallest such $k$ is $19$, and hence we must have $k = 19$ (because the next $k$ would be $19 + 37$, and the total amount will surely exceed $2023$).

We note that $2023 = 19 \times 87 + 370$. To pay exactly $2023$, we may start with $19$ coins with denomination $87$, and add $10$ units of $37$. This can be done by using a $74$ coin (which contributes $2$ units), a $111$ coin (which contributes $3$ units), or by replacing a $87$ coin by a $124$ coin (which contributes $1$ unit). Hence the problem is reduced to solving the equation $2a + 3b + c = 10$ in nonnegative integers. Since each choice of $(a, b)$ for which $2a + 3b \le 10$ gives a unique choice of $c$, we only have to solve the inequality $2a + 3b \le 10$ in nonnegative integers. We find that there are $14$ solutions as shown below, and so the answer is $14$.