By the QM-AM inequality note that \[ \sqrt{\frac{2a^2+3b^2}5}\ge\frac{2a+3b}5. \]Let the $2009$ positive integers be $a_1, a_2,\ldots, a_{2009}$ in clockwise order. Then, letting $a_0=a_{2009}$ and $a_{2010}=a_1$ we have that for all $1\le i\le 2009$, \[ a_i = \sqrt{\frac{2a_{i-1}^2+3a_{i+1}^2}5}\ge\frac{2a_{i-1}+3a_{i+1}}5, \]with equality if and only if $a_{i-1}=a_{i+1}$. Adding this for all $i$ we see that \[ a_1+a_2+\ldots+a_{2009}\ge\frac{2a_{2009}+3a_2}5+\frac{2a_1+3a_3}5+\ldots+\frac{2a_{2008}+3a_1}5=a_1+a_2+\ldots+a_{2009}. \]Since we achieve equality, we must have equality in all of our inequalities. Thus, the equality case is satisfied implying all $2009$ numbers must be equal. $\blacksquare$