Let $p_i$ for $i=1,2,..., k$ be a sequence of smallest consecutive prime numbers ($p_1=2$, $p_2=3$, $p_3=3$  etc. ). Let $N=p_1\cdot p_2 \cdot ... \cdot p_k$. Prove that in a set $\{ 1,2,...,N \}$ there exist exactly $\frac{N}{2}$ numbers which are divisible by odd number of primes $p_i$. 

[hide=example]For $k=2$ $p_1=2$, $p_2=3$, $N=6$. So in set $\{ 1,2,3,4,5,6 \}$ we can find $3$ number satisfying thesis: $2$, $3$ and $4$. ($1$ and $5$ are not divisible by $2$ or $3$, and $6$ is divisible by both of them so by even number of primes )[/hide]
1. **Define the problem and notation:**
   Let \( p_i \) for \( i=1,2,\ldots,k \) be the sequence of the smallest consecutive prime numbers. Specifically, \( p_1 = 2 \), \( p_2 = 3 \), \( p_3 = 5 \), etc. Define \( N \) as the product of these primes:
   \[
   N = p_1 \cdot p_2 \cdot \ldots \cdot p_k
   \]
   We need to prove that in the set \( \{1, 2, \ldots, N\} \), there are exactly \( \frac{N}{2} \) numbers which are divisible by an odd number of primes \( p_i \).

2. **Counting numbers divisible by an odd number of primes:**
   We can count the numbers divisible by an odd number of primes using the principle of inclusion-exclusion. Let \( \omega(n) \) denote the number of distinct prime divisors of \( n \). We need to count the numbers \( m \) such that \( \omega(m) \) is odd.

3. **Using the principle of inclusion-exclusion:**
   The principle of inclusion-exclusion can be used to count the numbers divisible by an odd number of primes. We start by considering the sum:
   \[
   \sum_{d \mid N, \omega(d) \equiv 1 \pmod{2}} \sum_{\substack{m \leq N \\ \gcd(m, N) = d}} 1
   \]
   This sum counts the numbers \( m \) in \( \{1, 2, \ldots, N\} \) that are divisible by an odd number of primes.

4. **Simplifying the inner sum:**
   The inner sum counts the numbers \( m \) such that \( \gcd(m, N) = d \). This is given by Euler's totient function \( \phi \):
   \[
   \sum_{\substack{m \leq N \\ \gcd(m, N) = d}} 1 = \phi\left(\frac{N}{d}\right)
   \]

5. **Summing over all divisors \( d \) with an odd number of prime factors:**
   \[
   \sum_{d \mid N, \omega(d) \equiv 1 \pmod{2}} \phi\left(\frac{N}{d}\right)
   \]

6. **Using generating functions:**
   Define the generating function \( f(x) \) as:
   \[
   f(x) := \prod_{i=1}^k (x + (p_i - 1))
   \]
   We want to calculate the sum of the coefficients corresponding to the monomial terms of degree \( k+1 \pmod{2} \).

7. **Evaluating the generating function:**
   The sum of the coefficients of the terms of degree \( k+1 \pmod{2} \) is given by:
   \[
   \frac{f(1) + (-1)^{k+1} f(-1)}{2}
   \]
   Since \( p_1 = 2 \), we have \( f(-1) = 0 \). Therefore:
   \[
   \frac{f(1) + (-1)^{k+1} \cdot 0}{2} = \frac{f(1)}{2}
   \]
   Since \( f(1) = p_1 p_2 \cdots p_k = N \), we get:
   \[
   \frac{N}{2}
   \]

8. **Conclusion:**
   Thus, there are exactly \( \frac{N}{2} \) numbers in the set \( \{1, 2, \ldots, N\} \) that are divisible by an odd number of primes \( p_i \).

\(\blacksquare\)

The final answer is \(\boxed{\frac{N}{2}}\)