Let $n$ be a natural number divisible by $4$. Determine the number of bijections $f$ on the set $\{1,2,...,n\}$ such that $f (j )+f^{-1}(j ) = n+1$ for $j = 1,..., n.$
1. Given that \( n \) is a natural number divisible by 4, we need to determine the number of bijections \( f \) on the set \(\{1, 2, \ldots, n\}\) such that \( f(j) + f^{-1}(j) = n + 1 \) for \( j = 1, \ldots, n \).

2. Let \( j = f(x) \). Then, we have \( f(f(x)) = n + 1 - x \). Notice that \( f(x) \neq x \) since \( n \) is even.

3. Define \( y = f(x) \). Applying \( f \) multiple times, we get a 4-number cycle: 
   \[
   x, y, n + 1 - x, n + 1 - y, x, y, n + 1 - x, n + 1 - y, \ldots
   \]
   In each cycle, the four numbers are different, and exactly two of them are in \(\{1, 2, \ldots, \frac{n}{2}\}\).

4. We need to count the number of ways to form these cycles. There are \((\frac{n}{2} - 1)(\frac{n}{2} - 3) \ldots 3 \cdot 1\) ways to group the pairs \((x, y)\).

5. Each of the \(\frac{n}{4}\) pairs can form 2 cycles: \((x, y, n + 1 - x, n + 1 - y)\) or \((x, n + 1 - y, n + 1 - x, y)\).

6. Therefore, the total number of bijections is:
   \[
   \left( (\frac{n}{2} - 1)(\frac{n}{2} - 3) \ldots 3 \cdot 1 \right) 2^{\frac{n}{4}}
   \]

7. We can simplify \((\frac{n}{2} - 1)(\frac{n}{2} - 3) \ldots 3 \cdot 1\) as:
   \[
   (\frac{n}{2} - 1)(\frac{n}{2} - 3) \ldots 3 \cdot 1 = \frac{(\frac{n}{2})!}{2^{\frac{n}{4}} (\frac{n}{4})!}
   \]

8. Thus, the total number of bijections is:
   \[
   \frac{(\frac{n}{2})!}{2^{\frac{n}{4}} (\frac{n}{4})!} \cdot 2^{\frac{n}{4}} = \frac{(\frac{n}{2})!}{(\frac{n}{4})!}
   \]

The final answer is \(\boxed{\frac{(\frac{n}{2})!}{(\frac{n}{4})!}}\).