Given that $5^{2018}$ has $1411$ digits and starts with $3$ (the leftmost non-zero digit is $3$), for how many integers $1\leq n\leq2017$ does $5^n$ start with $1$?

[i]2018 CCA Math Bonanza Tiebreaker Round #3[/i]
1. **Understanding the Problem:**
   We are given that \(5^{2018}\) has 1411 digits and starts with 3. We need to determine how many integers \(1 \leq n \leq 2017\) make \(5^n\) start with 1.

2. **Number of Digits in \(5^n\):**
   The number of digits \(d\) of a number \(x\) can be found using the formula:
   \[
   d = \lfloor \log_{10} x \rfloor + 1
   \]
   For \(5^n\), the number of digits is:
   \[
   d_n = \lfloor \log_{10} (5^n) \rfloor + 1 = \lfloor n \log_{10} 5 \rfloor + 1
   \]

3. **Condition for Starting with 1:**
   \(5^n\) starts with 1 if and only if \(1 \leq 5^n < 2\). Taking the logarithm base 10, we get:
   \[
   0 \leq \log_{10} (5^n) - k < \log_{10} 2 \quad \text{for some integer } k
   \]
   Simplifying, we have:
   \[
   k \leq n \log_{10} 5 < k + \log_{10} 2
   \]
   This can be rewritten as:
   \[
   \{ n \log_{10} 5 \} < \log_{10} 2
   \]
   where \(\{ x \}\) denotes the fractional part of \(x\).

4. **Number of Digits Change:**
   The number of digits of \(5^n\) and \(5^{n+1}\) changes when:
   \[
   \lfloor (n+1) \log_{10} 5 \rfloor \neq \lfloor n \log_{10} 5 \rfloor
   \]
   This happens when the fractional part \(\{ n \log_{10} 5 \}\) is close to 1. Specifically, it changes 1410 times as given in the problem.

5. **Counting \(n\) for \(5^n\) Starting with 1:**
   Since there are 2017 possible values for \(n\) and 1410 of them result in a change in the number of digits, the remaining values must be those where \(5^n\) starts with 1. Therefore, the number of such \(n\) is:
   \[
   2017 - 1410 = 607
   \]

The final answer is \(\boxed{607}\)