On Binary Island, residents communicate using special paper. Each piece of paper is a $1 \times n$ row of initially uncolored squares. To send a message, each square on the paper must either be colored either red or green. Unfortunately the paper on the island has become damaged, and each sheet of paper has $10$ random consecutive squares each of which is randomly colored red or green.

Malmer and Weven would like to develop a scheme that allows them to send messages of length $2016$ between one another. They would like to be able to send any message of length $2016$, and they want their scheme to work with perfect accuracy. What is the smallest value of $n$ for which they can develop such a strategy?

[i]Note that when sending a message, one can see which 10 squares are colored and what colors they are. One also knows on which square the message begins, and on which square the message ends.[/i]

1. **Claim**: The smallest value of \( n \) for which Malmer and Weven can develop a strategy to send messages of length 2016 with perfect accuracy is \( n = 2026 \).

2. **Proof of Minimum**:
    - Suppose \( n < 2026 \) works. It is clear \( n > 10 \) because 10 consecutive squares are randomly colored.
    - Consider a special paper whose first 10 squares are colored red. There are \( n - 10 \) squares remaining in the paper, each to be colored red or green, for a total of \( 2^{n-10} \) configurations.
    - However, there are \( 2^{2016} \) red-green messages of length 2016. By the Pigeonhole Principle, if \( 2^{n-10} < 2^{2016} \), then two of the messages, \( M_1 \) and \( M_2 \), must be represented by the same configuration \( C \).
    - Now consider what happens when Malmer sends \( C \) to Weven. Weven cannot decide if Malmer means \( M_1 \) or \( M_2 \), so the strategy fails.
    - Therefore, \( 2^{n-10} \geq 2^{2016} \) must hold, implying \( n - 10 \geq 2016 \) or \( n \geq 2026 \).

3. **Construction**:
    - Let red be \( 0 \) and green be \( 1 \). Then the messages and each sheet of paper can be encoded as a binary string.
    - If Malmer wants to send the binary string message \( M = a_1 a_2 \dots a_{2016} \) (\( a_i \in \{0, 1\} \)) on a paper \( P \), whose bits are \( p_1 p_2 p_3 \dots p_{2026} \) and \( p_i, p_{i+1}, p_{i+2}, \dots, p_{i+9} \) are fixed, then we will choose the \( p_i \) such that:
      \[
      \begin{aligned}
      p_1 + p_{11} &\equiv a_1 \pmod{2}, \\
      p_2 + p_{12} &\equiv a_2 \pmod{2}, \\
      p_3 + p_{13} &\equiv a_3 \pmod{2}, \\
      &\vdots \\
      p_i + p_{i+10} &\equiv a_i \pmod{2}, \\
      &\vdots \\
      p_{2016} + p_{2026} &\equiv a_{2016} \pmod{2}.
      \end{aligned}
      \]
    - This is well-defined because for a given \( 0 \le m \le 9 \), exactly one \( p_i \) with \( i \equiv m \pmod{10} \) is fixed, and we use the system of equations (in particular, \( p_{k-10} \equiv a_{k-10} - p_k \pmod{2} \), \( p_{k+10} \equiv a_k - p_k \pmod{2} \), and \( p_k \in \{0, 1\} \)) to define \( p_{i-10}, p_{i-20}, \dots \) and \( p_{i+10}, p_{i+20}, \dots \) uniquely.

4. **Verification**:
    - Suppose Weven receives the paper \( P \) from Malmer. He will read the message as:
      \[
      [(p_1 + p_{11}) \pmod{2}, (p_2 + p_{12}) \pmod{2}, (p_3 + p_{13}) \pmod{2}, \dots, (p_{2016} + p_{2026}) \pmod{2}].
      \]
    - This is exactly \( M \), so Malmer and Weven have a way to exchange messages.

5. **Conclusion**:
    - Having constructed \( n = 2026 \) and proven that \( n \le 2025 \) fails, we conclude that \( \boxed{2026} \) is the minimum \( n \) for which Malmer and Weven can develop a strategy.