For a sequence $(a_{n})_{n\geq 1}$ of real numbers it is known that $a_{n}=a_{n-1}+a_{n+2}$ for $n\geq 2$.
What is the largest number of its consecutive elements that can all be positive?
1. We start by analyzing the given recurrence relation for the sequence \((a_n)_{n \geq 1}\):
   \[
   a_n = a_{n-1} + a_{n+2} \quad \text{for} \quad n \geq 2.
   \]
   This implies that each term in the sequence is the sum of the previous term and the term two places ahead.

2. To find the maximum number of consecutive positive terms, we first consider a specific example. Let's take the sequence starting with \(a_1 = 1\), \(a_2 = 2\), \(a_3 = 3\), \(a_4 = 1\), and \(a_5 = 1\). We check if these terms satisfy the recurrence relation:
   \[
   a_2 = a_1 + a_4 \implies 2 = 1 + 1,
   \]
   \[
   a_3 = a_2 + a_5 \implies 3 = 2 + 1.
   \]
   Both equations are satisfied, and we have five consecutive positive terms: \(1, 2, 3, 1, 1\).

3. Next, we need to prove that it is impossible to have six consecutive positive terms. Assume there exists a sequence with six consecutive positive terms \(a_n, a_{n+1}, \ldots, a_{n+5}\). We write down the recurrence relations for these terms:
   \[
   a_{n+1} = a_n + a_{n+3},
   \]
   \[
   a_{n+2} = a_{n+1} + a_{n+4},
   \]
   \[
   a_{n+3} = a_{n+2} + a_{n+5}.
   \]

4. Adding these three equations together, we get:
   \[
   a_{n+1} + a_{n+2} + a_{n+3} = a_n + a_{n+3} + a_{n+1} + a_{n+4} + a_{n+2} + a_{n+5}.
   \]
   Simplifying, we cancel out the common terms on both sides:
   \[
   a_{n+2} + a_{n+3} = a_n + a_{n+4} + a_{n+5}.
   \]

5. Since \(a_n, a_{n+4},\) and \(a_{n+5}\) are all positive, the right-hand side of the equation is positive. However, the left-hand side \(a_{n+2} + a_{n+3}\) is also positive. This leads to a contradiction because the sum of positive terms cannot equal zero. Therefore, it is impossible to have six consecutive positive terms.

6. Hence, the maximum number of consecutive positive terms in the sequence is 5.

The final answer is \(\boxed{5}\).