68. National Mathematical Olympiad

If $|b|\le 2$, then $f(x)\ge 0$ for every $x$.

Let $|b|>2$ and $x_1<x_2$ are the real zeros of $f$. Then $$\begin{gathered} (\ast )f(f(x)+x)=(f(x)+x-x_1)(f(x)+x-x_2) \\ =(x-x_1)(x-x_2+1)(x-x_2)(x-x_1+1). \end{gathered}$$ We consider two cases.

Case 1. If $x_2-x_1\le 1$, then $2<|b|\le \sqrt{5}$. Since $f(\pm 1)=2\pm b$, it easily follows that the integers $x$ such that $f(f(x)+x)<0$ are exactly $-1$ and $-2$ when $b>0$, and $1$ and $0$ when $b<0$.

Case 2. If $x_2-x_1>1$, then $|b|>\sqrt{5}$. Now each of the intervals $(x_1-1,x_1)$ and $(x_2-1,x_2)$ contains exactly one integer $x$ such that $f(f(x)+x)<0$, unless $x_1$ or $x_2$ are not integers.

Summarizing, the required number is equal to 0 for $|b|\le 2$, to 1 for $b=m+1/m$, where $m\ge 2$ is integer, and to 2 in all other cases.