Determine all positive integers $n$ such that the decimal representation of the number $6^n + 1$ has all its digits the same.
1. **Initial Claim and Verification for Small Values of \( n \)**:
   - We claim that \( n = 1 \) and \( n = 5 \) are the only solutions.
   - Verify for \( n \leq 5 \):
     - For \( n = 1 \):
       \[
       6^1 + 1 = 6 + 1 = 7
       \]
       The number 7 has all its digits the same.
     - For \( n = 2 \):
       \[
       6^2 + 1 = 36 + 1 = 37
       \]
       The number 37 does not have all its digits the same.
     - For \( n = 3 \):
       \[
       6^3 + 1 = 216 + 1 = 217
       \]
       The number 217 does not have all its digits the same.
     - For \( n = 4 \):
       \[
       6^4 + 1 = 1296 + 1 = 1297
       \]
       The number 1297 does not have all its digits the same.
     - For \( n = 5 \):
       \[
       6^5 + 1 = 7776 + 1 = 7777
       \]
       The number 7777 has all its digits the same.

2. **Assume \( n \geq 6 \) and Analyze the Last Digit**:
   - Note that \( 6^n + 1 \equiv 1 \pmod{2} \) and \( 6^n + 1 \equiv 2 \pmod{5} \).
   - This implies the last digit of \( 6^n + 1 \) is 7.
   - Suppose \( 6^n + 1 = \overline{77\cdots 7} \) (a number with all digits 7).
   - Let \( N \) be the number of 7's in the decimal representation of \( 6^n + 1 \). Then \( N \geq 5 \).

3. **Formulate the Equation**:
   - The number \( \overline{77\cdots 7} \) with \( N \) digits can be written as:
     \[
     6^n + 1 = 7 \cdot \frac{10^N - 1}{9}
     \]
   - Multiply both sides by 9:
     \[
     9 \cdot 6^n + 9 = 7 \cdot (10^N - 1)
     \]
   - Simplify:
     \[
     9 \cdot 6^n + 9 = 7 \cdot 10^N - 7
     \]
     \[
     9 \cdot 6^n + 16 = 7 \cdot 10^N
     \]

4. **Analyze the Powers of 2**:
   - Let \( T = 9 \cdot 6^n + 16 \).
   - Clearly, \( v_2(T) = 4 \) because:
     \[
     9 \cdot 6^n = 9 \cdot (2 \cdot 3)^n = 9 \cdot 2^n \cdot 3^n
     \]
     - The term \( 9 \cdot 2^n \cdot 3^n \) has \( v_2(9 \cdot 2^n \cdot 3^n) = n \).
     - Adding 16 (which is \( 2^4 \)) to this term does not change the fact that the highest power of 2 in \( T \) is 4.
   - However, \( 7 \cdot 10^N \) has \( v_2(7 \cdot 10^N) = N \).

5. **Contradiction**:
   - Since \( N \geq 5 \), \( v_2(7 \cdot 10^N) \geq 5 \).
   - This contradicts \( v_2(T) = 4 \).

Therefore, the only possible values of \( n \) are those verified initially: \( n = 1 \) and \( n = 5 \).

The final answer is \( \boxed{ n = 1 } \) and \( n = 5 \).