We call a sequence of integers a [i]Fibonacci-type sequence[/i] if it is infinite in both ways and $a_{n}=a_{n-1}+a_{n-2}$ for any $n\in\mathbb{Z}$. How many [i]Fibonacci-type sequences[/i] can we find, with the property that in these sequences there are two consecutive terms, strictly positive, and less or equal than $N$ ? (two sequences are considered to be the same if they differ only by shifting of indices)

[i]Proposed by I. Pevzner[/i]
1. Let the two consecutive terms in the Fibonacci-type sequence be \(a_1\) and \(a_2\). We need to satisfy the conditions:
   \[
   a_2 \le a_1 \quad \text{and} \quad a_1 + a_2 \ge N + 1
   \]
   This is because \(a_2 - a_1 \le 0\) implies \(a_2 \le a_1\), and \(a_1 + a_2 > N\) implies \(a_1 + a_2 \ge N + 1\).

2. We need to count the number of pairs \((a_1, a_2)\) that satisfy these conditions.

### Case 1: \(N\) is even
3. Since \(a_1 \ge a_2\) and \(a_1 + a_2 \ge N + 1\), we have:
   \[
   2a_1 \ge N + 1 \implies a_1 \ge \frac{N}{2} + \frac{1}{2}
   \]
   Since \(a_1\) must be an integer, we have:
   \[
   a_1 \ge \frac{N}{2} + 1
   \]
   Let \(a_1 = \frac{N}{2} + k\), where \(k\) is a positive integer such that \(1 \le k \le \frac{N}{2}\) (since \(a_1 \le N\)).

4. Similarly, we can write \(a_2 = \frac{N}{2} + p\), where \(p\) is an integer such that:
   \[
   -\frac{N}{2} + 1 \le p \le \frac{N}{2} \quad \text{and} \quad p \le k
   \]
   This is because \(a_1 \ge a_2\) and \(0 < a_2 \le N\).

5. For each value of \(k\), \(1 \le k \le \frac{N}{2}\), there are \(2k\) values of \(p\). Therefore, the total number of pairs \((a_1, a_2)\) is:
   \[
   \sum_{i=1}^{\frac{N}{2}} 2i = 2 \sum_{i=1}^{\frac{N}{2}} i = 2 \cdot \frac{\frac{N}{2} (\frac{N}{2} + 1)}{2} = \left(\frac{N}{2}\right) \left(\frac{N}{2} + 1\right)
   \]

### Case 2: \(N\) is odd
6. Since \(a_1 \ge a_2\) and \(a_1 + a_2 \ge N + 1\), we have:
   \[
   2a_1 \ge N + 1 \implies a_1 \ge \frac{N + 1}{2}
   \]
   Let \(a_1 = \frac{N - 1}{2} + k\), where \(k\) is a positive integer such that \(1 \le k \le \frac{N + 1}{2}\) (since \(a_1 \le N\)).

7. Similarly, we can write \(a_2 = \frac{N - 1}{2} + p\), where \(p\) is an integer such that:
   \[
   -\frac{N - 1}{2} + 1 \le p \le \frac{N + 1}{2} \quad \text{and} \quad p \le k
   \]
   This is because \(a_1 \ge a_2\) and \(0 < a_2 \le N\).

8. For each value of \(k\), \(1 \le k \le \frac{N + 1}{2}\), there are \(2k - 1\) values of \(p\). Therefore, the total number of pairs \((a_1, a_2)\) is:
   \[
   \sum_{i=1}^{\frac{N + 1}{2}} (2i - 1) = \left(\frac{N + 1}{2}\right)^2
   \]

### Conclusion
- If \(N\) is even, the number of Fibonacci-type sequences is:
  \[
  \left(\frac{N}{2}\right) \left(\frac{N}{2} + 1\right)
  \]
- If \(N\) is odd, the number of Fibonacci-type sequences is:
  \[
  \left(\frac{N + 1}{2}\right)^2
  \]

The final answer is \( \boxed{ \left(\frac{N}{2}\right) \left(\frac{N}{2} + 1\right) } \) if \(N\) is even and \(\left(\frac{N + 1}{2}\right)^2\) if \(N\) is odd.