Answer: $k \in 1, 2, 4$

Suppose that $k \ge 5$. It is clear that among any $k$ consecutive numbers, there is one that is divisible by $5$, and one that is divisible by $2$, so their product ends in $0$, hence $k$ is divisible by $10$. It is clear that then $k \ge 10$, so in the product of $k$ consecutive numbers there are at least two numbers divisible by $5$ and at least two numbers divisible by $2$, and one of them is divisible by $k$. Hence, the product is divisible by $10k$, so there will be more zeros at the end than zeros at the end of $k$, which is where we get the contradiction. Thus, $k \le 4$.

If $k = 3$, then the product will be even, but it cannot end in the digit $3$.

For $k = 1, 2, 4$ it is enough to consider the following examples: $1$, $1 \cdot 2 = 2$, $1 \cdot 2 \cdot 3 \cdot 4 = 24$.