Let $n $ be a positive integer. Let $f $ be a function from nonnegative integers to themselves. Let $f (0,i)=f (i,0)=0$, $f (1, 1)=n $, and $ f(i, j)= [\frac {f(i-1,j)}{2}]+ [\frac {f(i, j-1)}{2}] $ for positive integers $i, j$ such that $i*j>1$. Find the number of pairs $(i,j) $ such that $f (i, j) $ is an odd number.( $[x]$ is the floor function).
1. **Define the function and initial conditions:**
   Given the function \( f \) from nonnegative integers to themselves, we have:
   \[
   f(0, i) = f(i, 0) = 0, \quad f(1, 1) = n
   \]
   and for positive integers \( i, j \) such that \( i \cdot j > 1 \):
   \[
   f(i, j) = \left\lfloor \frac{f(i-1, j)}{2} \right\rfloor + \left\lfloor \frac{f(i, j-1)}{2} \right\rfloor
   \]

2. **Define the sum of the numbers on the \( k \)-th diagonal:**
   Let \( g(k) \) be the sum of the numbers on the \( k \)-th diagonal:
   \[
   g(k) = f(k, 0) + f(k-1, 1) + f(k-2, 2) + \cdots + f(1, k-1) + f(0, k)
   \]

3. **Express \( g(k+1) \) in terms of \( g(k) \):**
   Note that:
   \[
   g(k+1) = f(k+1, 0) + f(0, k+1) + \sum_{i=1}^{k} f(k+1-i, i)
   \]
   Using the recurrence relation for \( f \):
   \[
   f(k+1-i, i) = \left\lfloor \frac{f(k-i, i)}{2} \right\rfloor + \left\lfloor \frac{f(k+1-i, i-1)}{2} \right\rfloor
   \]
   Therefore:
   \[
   g(k+1) = \sum_{i=1}^{k} \left( \left\lfloor \frac{f(k-i, i)}{2} \right\rfloor + \left\lfloor \frac{f(k+1-i, i-1)}{2} \right\rfloor \right)
   \]

4. **Simplify the expression for \( g(k+1) \):**
   Since each term in the sum is halved and floored:
   \[
   g(k+1) = \sum_{i=1}^{k} \left( \left\lfloor \frac{f(k-i, i)}{2} \right\rfloor + \left\lfloor \frac{f(k+1-i, i-1)}{2} \right\rfloor \right)
   \]
   This can be simplified to:
   \[
   g(k+1) = \sum_{i=1}^{k-1} 2 \left\lfloor \frac{f(k-i, i)}{2} \right\rfloor
   \]

5. **Determine the number of odd entries:**
   Using the property \( x = 2 \left\lfloor \frac{x}{2} \right\rfloor + \delta_x \), where \( \delta_x = 0 \) if \( x \) is even and \( \delta_x = 1 \) if \( x \) is odd, we infer that \( g(k) - g(k+1) \) gives the number of odd integers on the \( k \)-th diagonal.

6. **Find the total number of odd integers:**
   Since \( g(k) \) is a decreasing sequence of nonnegative integers, it eventually becomes constant. This implies that there exists an integer \( t \) such that \( g(t) = g(t+1) = g(t+2) = \cdots \), meaning there are no odd entries on the \( k \)-th diagonal for \( k \ge t \).

7. **Sum the differences to find the total number of odd integers:**
   The total number of odd integers is:
   \[
   \sum_{k=1}^{p-1} (g(k) - g(k+1)) = g(1) - g(p)
   \]
   Since \( g(p) = 0 \), we have:
   \[
   g(1) - g(p) = f(1, 1) = n
   \]

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