For each positive integer $n$, let $s(n)$ be the sum of the digits of $n$. Find the smallest positive integer $k$ such that
\[s(k) = s(2k) = s(3k) = \cdots = s(2013k) = s(2014k).\]
To find the smallest positive integer \( k \) such that 
\[ s(k) = s(2k) = s(3k) = \cdots = s(2013k) = s(2014k), \]
we need to analyze the sum of the digits function \( s(n) \) and how it behaves under multiplication.

1. **Initial Observations**:
   - If \( k \) is a single-digit number, then \( s(11k) = 2s(k) \), which is a contradiction because \( s(11k) \) would be the sum of the digits of a two-digit number.
   - If \( k \) has two digits, then \( s(101k) = 2s(k) \), which is again a contradiction because \( s(101k) \) would be the sum of the digits of a three-digit number.
   - If \( k \) has three digits, then \( s(1001k) = 2s(k) \), which is also a contradiction because \( s(1001k) \) would be the sum of the digits of a four-digit number.

2. **Considering Four-Digit \( k \)**:
   - Let \( k = a_3a_2a_1a_0 \) (where \( a_i \) are the digits of \( k \)).
   - We need \( s(1001k) = s(k) \). Writing \( 1001k \) as \( a_3a_2a_1a_0 \cdot 1001 \), we get:
     \[
     1001k = a_3a_2a_1a_0 \cdot 1001 = a_3a_2a_1a_0 + 0a_3a_2a_1a_0
     \]
     This implies that the sum of the digits of \( 1001k \) must be equal to the sum of the digits of \( k \).

3. **Analyzing the Sum of Digits**:
   - For \( k = 9999 \):
     \[
     s(9999) = 9 + 9 + 9 + 9 = 36
     \]
   - For any \( a \) with at most four digits, \( 9999a \) can be written as:
     \[
     9999a = a_3a_2a_1a_0 \cdot 9999 = a_3a_2a_1a_0 \cdot (10000 - 1) = a_3a_2a_1a_0 \cdot 10000 - a_3a_2a_1a_0
     \]
     This results in a number where the sum of the digits is:
     \[
     s(9999a) = 36
     \]
     for any \( a \) with at most four digits.

4. **Verification**:
   - For \( k = 9999 \), we need to show that \( s(ak) = 36 \) for all \( a = 1, 2, \ldots, 2014 \).
   - If \( a = a_3a_2a_1a_0 \) (with \( a_3 \) possibly zero), then:
     \[
     9999a = a_3a_2a_1a_0 \cdot 9999 = a_3a_2a_1a_0 \cdot (10000 - 1)
     \]
     This results in:
     \[
     s(9999a) = 36
     \]
     for all \( a \) with at most four digits.

Thus, the smallest positive integer \( k \) such that \( s(k) = s(2k) = \cdots = s(2014k) \) is \( k = 9999 \).

The final answer is \( \boxed{9999} \).