1. Define the problem and notation:
- We have $n$ rooms numbered $1, 2, \ldots, n$.
- Among these rooms, $k$ rooms are under repair.
- 100 tourists arrive and need to find a free room without checking an occupied room.
- We need to find the smallest $n$ such that the tourists can always find a free room without checking an occupied room.

2. **Define the function $f(i, j)$:**
- Let $f(i, j)$ be the $j$-th room that the $i$-th tourist will check in the strategy.
- The tourists need to coordinate their strategy to avoid checking an occupied room.

3. Lemma:
- If $f(i_1, j_1) = f(i_2, j_2)$ for some $i_1 < i_2$, then $j_1 + j_2 \geq k + 3$.
- This lemma ensures that when the $i_2$-th tourist is checking the $j_2$-th room, they won't check an occupied room.

4. Proof of the Lemma:
- $(\Rightarrow)$ If $f(i_1, j_1)$ is under repair, then the lemma holds.
- If $i_2$ had checked another room that is occupied by $i_1$, then $i_2$ would have exploded.
- Otherwise, in the sequence $f(i_1, 1), f(i_1, 2), \ldots, f(i_1, j_1-1), f(i_2, 1), f(i_2, 2), \ldots, f(i_2, j_2-1)$, there are at least $k+1$ rooms, and therefore there exists at least one room that is not under repair.
- Hence, either $f(i_1, j_1)$ won't be occupied by $i_1$, or $i_2$ won't check this room.
- $(\Leftarrow)$ If $j_1 + j_2 \leq k + 2$, and the $k$ rooms $f(i_1, 1), f(i_1, 2), \ldots, f(i_1, j_1-1), f(i_2, 1), f(i_2, 2), \ldots, f(i_2, j_2-1)$ are under repair, then $i_2$ will explode in this strategy when checking the $j_2$-th room.

5. Strategy Success Condition:
- A strategy will succeed if and only if for all $i_1 \neq i_2$ and $j_1, j_2$, if $f(i_1, j_1) = f(i_2, j_2)$, then $j_1 + j_2 \geq k + 3$.

6. Case Analysis:
- If $k = 2l - 1$ (odd), then for all $i_1 \neq i_2$, $(j_1, j_2 \leq l$ or $j_1 = l + 1, j_2 \leq l)$, $f(i_1, j_1) \neq f(i_2, j_2)$. Therefore, $n \geq 100l + 1$.
- If $k = 2l$ (even), then for all $i_1 \neq i_2$, $j_1, j_2 \leq l + 1$, $f(i_1, j_1) \neq f(i_2, j_2)$. Therefore, $n \geq 100(l + 1)$.

7. Construction of the Strategy:
- If $k = 2l - 1$ (odd), let:
$$f(i, j) = \begin{cases} l(i-1) + j & \text{if } j \leq l \\ 100l + 1 & \text{if } j = l + 1 \\ l(100 - i) + 2l + 2 - j & \text{if } j \geq l + 2 \end{cases}$$
- If $k = 2l$ (even), let:
$$f(i, j) = \begin{cases} (l + 1)(i-1) + j & \text{if } j \leq l + 1 \\ l(100 - i) + 2l + 3 - j & \text{if } j \geq l + 2 \end{cases}$$

8. Conclusion:
- For $k$ odd: $n \geq 50k + 51$.
- For $k$ even: $n \geq 50k + 100$.

The final answer is $2 \nmid k: 50k + 51$ and $2 \mid k: 50k + 100$.