= Solution
First establish a uniform upper bound. Direct subtraction gives
$$
F(N)-\frac32=-\frac N2(2N^2-7N+7)<0\qquad(N>0),
$$
because the <quadratic polynomial> has positive leading coefficient and <discriminant> $49-56=-7$. The assumed positivity of $F(N_0)$ therefore places $N_1$ in $(0,3/2)$.
This interval is <forward invariant>. On $[0,1/2]$ and on $[1,3/2]$, the three factors in $P(N)$ have nonnegative product, hence $F(N)\ge1/2$. On $[1/2,1]$ the product is nonpositive, but
$$
|P(N)|=(N-\tfrac12)(1-N)(2-N)\le\frac12\frac12\frac32=\frac38,
$$
so $F(N)\ge1/8>0$. Together with the upper bound this proves $F((0,3/2))\subset(0,3/2)$. Induction now gives
$$
\boxed{N_t>0\quad\text{for every }t\ge0.}
$$
The hypothesis on the first image is essential for initial values outside that <invariant interval>.
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