We claim that the maximum is 5. Consider a sequence that begins with 1, 2, 3, 1, 1. In this sequence, the equations $a_{2}=a_{1}+a_{4}$ and $a_{3}=a_{2}+a_{5}$ are both satisfied, and we have five consecutive positive terms. Now suppose there is such a sequence with six consecutive positive terms $a_{n},a_{n+1}\ldots ,a_{n+5}$. We have the following equations: $a_{n+1}=a_{n}+a_{n+3}$ $a_{n+2}=a_{n+1}+a_{n+4}$ $a_{n+3}=a_{n+2}+a_{n+5}$ Adding together these three equations and cancelling terms that appear on both sides, we get $0=a_{n}+a_{n+4}+a_{n+5}$, a contradiction since $a_{n}$, $a_{n+4}$, and $a_{n+5}$ are all positive. Therefore, the maximum number of consecutive terms is indeed 5.