[sp]For the whole sum to be equal to $1$, the sum of the positive terms must be almost the same as the sum of the negative terms. In order to minimise the number of positive terms, it seems clear that we should take the positive terms to be as large as possible (so that there will be fewest of them).
Writing $n$ for $2009$, the sum of all the numbers from $1$ to $n$ is $\frac12n(n+1).$ Suppose that there is a number $r < n$ such that the sum of the $r$ largest numbers in the sequence (namely the numbers $n-r+1$ to $n$ inclusive) is exactly half of the total. The condition for that is $$rn - \tfrac12r(r-1) = \tfrac14n(n+1).$$ Solve that quadratic equation for $r$ to get $$r = \tfrac12\bigl(2n+1 - \sqrt{2n^2 + 2n + 1}\bigr)$$ (taking the negative sign for the square root to ensure that $r<n$). Plugging in the value $n=2009$, I get $r = 588.56...$.
So it looks as though we should take the $588$ largest numbers on the sequence (namely $1422$ to $2009$ inclusive) to have positive signs, and then to have just one further positive term to make the sum of the positive terms $1$ more than the sum of the remaining (negative) terms.
In fact, the sum of all the numbers from $1$ to $2009$ is $2019045$. Half of that is $1009522.5$. So we want the sum of the positive numbers in the sequence to be $1009523$, and the sum of the negative numbers in the sequence to be $1009522.$ The first term in the sum (namely $1$) has no $\triangle$ in front of it, so must necessarily be positive. The sum of the numbers $1422$ to $2009$ is $1008714$, leaving a shortfall of $1009523 - 1008715 = 808$ for the sum of the positive terms.
Thus the minimal number of positive terms is given by taking the numbers $808$, and $1422$ to $2009$ inclusive, to be preceded by $\triangle = +$, and all the remaining triangles to be $\triangle = -.$ That gives the minimal number of "$+$" triangles to be $589.$[/sp]
