Let $f:\mathbb{N}\to\mathbb{N}$ be a function satisfying the following conditions:

(a) $f(1)=1$.
(b) $f(a)\leq f(b)$ whenever $a$ and $b$ are positive integers with $a\leq b$.
(c) $f(2a)=f(a)+1$ for all positive integers $a$.

How many possible values can the $2014$-tuple $(f(1),f(2),\ldots,f(2014))$ take?
To solve this problem, we need to analyze the function \( f \) based on the given conditions and determine the number of possible values for the 2014-tuple \((f(1), f(2), \ldots, f(2014))\).

1. **Initial Condition**:
   \[
   f(1) = 1
   \]

2. **Monotonicity**:
   \[
   f(a) \leq f(b) \quad \text{whenever} \quad a \leq b
   \]
   This implies that \( f \) is a non-decreasing function.

3. **Functional Equation**:
   \[
   f(2a) = f(a) + 1 \quad \text{for all positive integers} \quad a
   \]

Let's use these conditions to determine the values of \( f \) for the first few integers and identify a pattern.

4. **Calculate \( f \) for the first few integers**:
   \[
   \begin{aligned}
   &f(1) = 1 \\
   &f(2) = f(2 \cdot 1) = f(1) + 1 = 2 \\
   &f(4) = f(2 \cdot 2) = f(2) + 1 = 3 \\
   &f(8) = f(2 \cdot 4) = f(4) + 1 = 4 \\
   &\vdots \\
   &f(2^k) = k + 1 \quad \text{for} \quad k \in \mathbb{N}
   \end{aligned}
   \]

5. **Determine \( f \) for other integers**:
   Since \( f \) is non-decreasing and \( f(2a) = f(a) + 1 \), we can infer that for any integer \( n \), \( f(n) \) is determined by the largest power of 2 less than or equal to \( n \).

   For example:
   \[
   \begin{aligned}
   &f(3) = f(2) = 2 \quad (\text{since } 2 \leq 3 < 4) \\
   &f(5) = f(4) = 3 \quad (\text{since } 4 \leq 5 < 8) \\
   &f(6) = f(4) = 3 \quad (\text{since } 4 \leq 6 < 8) \\
   &f(7) = f(4) = 3 \quad (\text{since } 4 \leq 7 < 8) \\
   &f(9) = f(8) = 4 \quad (\text{since } 8 \leq 9 < 16) \\
   &\vdots
   \end{aligned}
   \]

6. **General Pattern**:
   For any integer \( n \), \( f(n) \) is determined by the largest power of 2 less than or equal to \( n \). Specifically, if \( 2^k \leq n < 2^{k+1} \), then \( f(n) = k + 1 \).

7. **Determine the number of distinct values in the 2014-tuple**:
   We need to find the range of \( f(n) \) for \( n \) from 1 to 2014. The largest power of 2 less than or equal to 2014 is \( 2^{10} = 1024 \), and the next power of 2 is \( 2^{11} = 2048 \).

   Therefore, \( f(n) \) ranges from \( f(1) = 1 \) to \( f(1024) = 11 \). Since \( f \) is non-decreasing and \( f(2^k) = k + 1 \), the values of \( f(n) \) for \( n \) from 1 to 2014 are \( 1, 2, 3, \ldots, 11 \).

   Hence, there are 11 distinct values in the 2014-tuple.

The final answer is \(\boxed{11}\).