Juan writes the list of pairs $(n, 3^n)$, with $n=1, 2, 3,...$ on a chalkboard. As he writes the list, he underlines the pairs $(n, 3^n)$ when $n$ and $3^n$ have the same units digit. What is the $2013^{th}$ underlined pair?
1. **Identify the units digits of \(3^n\):**
   The units digit of \(3^n\) follows a repeating cycle of 4: 
   \[
   \begin{aligned}
   3^1 &\equiv 3 \pmod{10}, \\
   3^2 &\equiv 9 \pmod{10}, \\
   3^3 &\equiv 7 \pmod{10}, \\
   3^4 &\equiv 1 \pmod{10}, \\
   3^5 &\equiv 3 \pmod{10}, \\
   &\vdots
   \end{aligned}
   \]
   Therefore, the units digits of \(3^n\) are 3, 9, 7, 1, and then the cycle repeats.

2. **Determine when \(n\) and \(3^n\) have the same units digit:**
   - For \(n \equiv 1 \pmod{4}\), \(3^n \equiv 3 \pmod{10}\).
   - For \(n \equiv 3 \pmod{4}\), \(3^n \equiv 7 \pmod{10}\).

3. **Find the sequence of \(n\) values that satisfy the conditions:**
   - If \(n \equiv 1 \pmod{4}\), then \(n\) can be written as \(n = 4k + 1\).
   - If \(n \equiv 3 \pmod{4}\), then \(n\) can be written as \(n = 4k + 3\).

4. **Combine the sequences:**
   - The sequence of \(n\) values that satisfy the conditions is \(1, 3, 5, 7, 9, 11, 13, 15, \ldots\).

5. **Determine the 2013th underlined pair:**
   - The sequence of \(n\) values is an arithmetic sequence with the first term \(a = 1\) and common difference \(d = 2\).
   - The \(k\)-th term of this sequence is given by:
     \[
     n_k = 1 + (k-1) \cdot 2 = 2k - 1
     \]
   - For the 2013th underlined pair, we set \(k = 2013\):
     \[
     n_{2013} = 2 \cdot 2013 - 1 = 4025
     \]

6. **Conclusion:**
   - The 2013th underlined pair is \((4025, 3^{4025})\).

The final answer is \(\boxed{(4025, 3^{4025})}\).