Determine the number of all positive integers which cannot be written in the form $80k + 3m$, where $k,m \in N = \{0,1,2,...,\}$
1. **Identify the problem and the form of the numbers:**
   We need to determine the number of positive integers that cannot be written in the form \(80k + 3m\), where \(k, m \in \mathbb{N} = \{0, 1, 2, \ldots\}\).

2. **Use the known result from the Frobenius coin problem:**
   According to the Frobenius coin problem, for two coprime integers \(a\) and \(b\), the largest number that cannot be expressed as \(ax + by\) (where \(x, y \in \mathbb{N}\)) is \(ab - a - b\). Here, \(a = 80\) and \(b = 3\), which are coprime.

   Therefore, the largest number that cannot be written as \(80k + 3m\) is:
   \[
   80 \cdot 3 - 80 - 3 = 240 - 80 - 3 = 157
   \]

3. **Identify the range of numbers to check:**
   All numbers greater than 157 can be written in the form \(80k + 3m\). We need to check the numbers from 0 to 157 to find the ones that cannot be written in this form.

4. **Check numbers for \(k = 0\):**
   When \(k = 0\), the equation becomes \(3m = x\). Thus, \(x\) must be a multiple of 3. Therefore, if \(x\) is a multiple of 3, it is a good number.

5. **Check numbers for \(k = 1\):**
   When \(k = 1\), the equation becomes \(80 + 3m = x\). Rearranging, we get \(3m = x - 80\). Thus, \(x - 80\) must be a multiple of 3. Therefore, if \(x \equiv 1 \pmod{3}\) and \(80 \leq x \leq 157\), then \(x\) is a bad number.

6. **Identify bad numbers in the range \(80 \leq x \leq 157\):**
   We need to find numbers \(x\) in this range such that \(x \equiv 1 \pmod{3}\):
   \[
   80 \leq x \leq 157 \quad \text{and} \quad x \equiv 1 \pmod{3}
   \]
   The sequence of such numbers is:
   \[
   82, 85, 88, \ldots, 157
   \]
   This is an arithmetic sequence with the first term \(a = 82\) and common difference \(d = 3\). The number of terms \(n\) in this sequence can be found using the formula for the \(n\)-th term of an arithmetic sequence:
   \[
   a_n = a + (n-1)d
   \]
   Setting \(a_n = 157\):
   \[
   157 = 82 + (n-1) \cdot 3 \implies 157 = 82 + 3n - 3 \implies 157 = 79 + 3n \implies 3n = 78 \implies n = 26
   \]
   So, there are 26 bad numbers in this range.

7. **Identify bad numbers in the range \(0 \leq x < 80\):**
   We need to find numbers \(x\) in this range such that \(x \equiv 1 \pmod{3}\) or \(x \equiv 2 \pmod{3}\):
   \[
   1, 2, 4, 5, 7, 8, \ldots, 79
   \]
   This sequence consists of two interleaved arithmetic sequences:
   - \(1, 4, 7, \ldots, 79\) (numbers \(x \equiv 1 \pmod{3}\))
   - \(2, 5, 8, \ldots, 77\) (numbers \(x \equiv 2 \pmod{3}\))

   Each sequence has a common difference of 3. The number of terms in each sequence can be found similarly:
   - For \(1, 4, 7, \ldots, 79\):
     \[
     a_n = 1 + (n-1) \cdot 3 \implies 79 = 1 + 3(n-1) \implies 79 = 3n - 2 \implies 3n = 81 \implies n = 27
     \]
   - For \(2, 5, 8, \ldots, 77\):
     \[
     a_n = 2 + (n-1) \cdot 3 \implies 77 = 2 + 3(n-1) \implies 77 = 3n - 1 \implies 3n = 78 \implies n = 26
     \]

   So, there are 27 bad numbers in the first sequence and 26 bad numbers in the second sequence, making a total of:
   \[
   27 + 26 = 53
   \]

8. **Combine the results:**
   The total number of bad numbers is:
   \[
   26 \text{ (from } 80 \leq x \leq 157) + 53 \text{ (from } 0 \leq x < 80) = 79
   \]

The final answer is \(\boxed{79}\)