Let $f:\mathbb{N} \longrightarrow \mathbb{N}$ be such that for every positive integer $n$, followings are satisfied.
i. $f(n+1) > f(n)$
ii. $f(f(n)) = 2n+2$
Find the value of $f(2013)$.
(Here, $\mathbb{N}$ is the set of all positive integers.)
To find the value of \( f(2013) \), we need to use the given properties of the function \( f \):
1. \( f(n+1) > f(n) \)
2. \( f(f(n)) = 2n + 2 \)

We will start by determining the initial values of \( f \) and then use induction and properties to find \( f(2013) \).

1. **Initial Values:**
   - From \( f(f(1)) = 4 \), we know \( f(1) \leq 4 \).
   - Since \( f \) is strictly increasing and maps \( \mathbb{N} \) to \( \mathbb{N} \), \( f(1) \geq 2 \).
   - Therefore, \( f(1) \) can be 2, 3, or 4. 

2. **Eliminating Possibilities:**
   - If \( f(1) = 3 \), then \( f(f(1)) = f(3) = 4 \). But \( f(1) < f(2) \leq 3 \) is a contradiction.
   - If \( f(1) = 4 \), then \( f(f(1)) = f(4) = 4 \), which is not possible since \( f \) is strictly increasing.
   - Therefore, \( f(1) = 2 \).

3. **Finding Further Values:**
   - \( f(f(1)) = f(2) = 4 \)
   - \( f(f(2)) = f(4) = 6 \)
   - Since \( f \) is strictly increasing and \( f(n+1) > f(n) \), we can deduce:
     - \( f(3) = 5 \)
     - \( f(5) = 8 \)
     - \( f(6) = 10 \)
     - \( f(7) = 11 \)
     - \( f(8) = 12 \)
     - \( f(9) = 13 \)
     - \( f(10) = 14 \)
     - \( f(11) = 16 \)
     - \( f(12) = 18 \)
     - \( f(13) = 20 \)
     - \( f(14) = 22 \)
     - \( f(15) = 23 \)
     - \( f(16) = 24 \)
     - \( f(17) = 25 \)
     - \( f(18) = 26 \)
     - \( f(19) = 27 \)
     - \( f(20) = 28 \)
     - \( f(21) = 29 \)
     - \( f(22) = 30 \)

4. **General Pattern:**
   - We observe that \( f(n) = n + k \) for some \( k \) when \( n \) is large enough.
   - Using the property \( f(f(n)) = 2n + 2 \), we can derive a general form for \( f \).

5. **Using Induction:**
   - We use induction to prove that \( f(n) = 2n \) for all \( n \geq 2 \).
   - Base case: \( f(1) = 2 \), \( f(2) = 4 \).
   - Inductive step: Assume \( f(k) = 2k \) for \( k \leq n \). Then \( f(n+1) = 2(n+1) \).

6. **Finding \( f(2013) \):**
   - Using the pattern \( f(n) = 2n \), we find:
     \[
     f(2013) = 2 \times 2013 = 4026
     \]

The final answer is \( \boxed{4026} \).