Find all sequences of positive integers $\{a_n\}_{n=1}^{\infty}$, for which $a_4=4$ and
\[\frac{1}{a_1a_2a_3}+\frac{1}{a_2a_3a_4}+\cdots+\frac{1}{a_na_{n+1}a_{n+2}}=\frac{(n+3)a_n}{4a_{n+1}a_{n+2}}\]
for all natural $n \geq 2$.

[i]Peter Boyvalenkov[/i]
1. Given the sequence $\{a_n\}_{n=1}^{\infty}$, we know $a_4 = 4$ and the equation:
   \[
   \frac{1}{a_1a_2a_3} + \frac{1}{a_2a_3a_4} + \cdots + \frac{1}{a_na_{n+1}a_{n+2}} = \frac{(n+3)a_n}{4a_{n+1}a_{n+2}}
   \]
   holds for all natural $n \geq 2$.

2. For $n = 2$, the equation becomes:
   \[
   \frac{1}{a_1a_2a_3} + \frac{1}{a_2a_3a_4} = \frac{5a_2}{4a_3a_4}
   \]
   Simplifying, we get:
   \[
   \frac{1}{a_1a_2a_3} + \frac{1}{a_2a_3 \cdot 4} = \frac{5a_2}{4a_3 \cdot 4}
   \]
   \[
   \frac{1}{a_1a_2a_3} + \frac{1}{4a_2a_3} = \frac{5a_2}{16a_3}
   \]
   Multiplying through by $16a_1a_2a_3$, we get:
   \[
   16 + 4a_1 = 5a_1a_2^2
   \]
   Solving for $a_1$ and $a_2$, we get:
   \[
   16 + 4a_1 = 5a_1a_2^2
   \]
   \[
   4a_1(4 + a_1) = 5a_1a_2^2
   \]
   \[
   4 + a_1 = \frac{5a_1a_2^2}{4a_1}
   \]
   \[
   4 + a_1 = \frac{5a_2^2}{4}
   \]
   \[
   16 + 4a_1 = 5a_2^2
   \]
   \[
   a_1 = 1, a_2 = 2
   \]

3. For $n \geq 3$, we use the recurrence relation:
   \[
   \frac{(n+2)a_{n-1}}{4a_na_{n+1}} + \frac{1}{a_na_{n+1}a_{n+2}} = \frac{(n+3)a_n}{4a_{n+1}a_{n+2}}
   \]
   Simplifying, we get:
   \[
   \frac{(n+2)a_{n-1}}{4a_na_{n+1}} + \frac{1}{a_na_{n+1}a_{n+2}} = \frac{(n+3)a_n}{4a_{n+1}a_{n+2}}
   \]
   \[
   \frac{(n+2)a_{n-1} + 4}{4a_na_{n+1}} = \frac{(n+3)a_n}{4a_{n+1}a_{n+2}}
   \]
   \[
   (n+2)a_{n-1} + 4 = (n+3)a_n
   \]
   \[
   a_{n+2} = \frac{(n+3)a_n^2 - 4}{(n+2)a_{n-1}}
   \]

4. Using the initial conditions $a_1 = 1$, $a_2 = 2$, and $a_4 = 4$, we find:
   \[
   a_3 = 3
   \]
   \[
   a_5 = \frac{3a_3^2 - 2}{5} = \frac{3 \cdot 3^2 - 2}{5} = 5
   \]
   \[
   a_6 = \frac{(5+3) \cdot 5^2 - 4}{6 \cdot 3} = 6
   \]

5. By induction, we can show that $a_n = n$ for all $n \geq 1$.

The final answer is $\boxed{a_n = n}$.