No. The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials

$$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!}$$

and as such if

$$\displaystyle f(x) = \sum a_n {x \choose n}, a_n \in \mathbb{Z}$$

is an integer-valued polynomial, then, for example, $f \left( - \frac{1}{2} \right)$ must have denominator a power of $2$, since

$$\displaystyle {-\frac{1}{2} \choose n} = \frac{(-1)(-3)\dots(-2n+1)}{n!} = (-1)^n \frac{ {2n \choose n} }{2^n}$$

and hence no such polynomial $f$ can pass through a point like $\left( - \frac{1}{2}, \frac{1}{3} \right)$.

No. The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials

$$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!}$$

and as such if

$$\displaystyle f(x) = \sum a_n {x \choose n}, a_n \in \mathbb{Z}$$

is an integer-valued polynomial, then, for example, $f \left( - \frac{1}{2} \right)$ must have denominator a power of $2$, since

$$\displaystyle {-\frac{1}{2} \choose n} = \frac{(-1)(-3)\dots(-2n+1)}{2^n n!} = (-1)^n \frac{ {2n \choose n} }{2^n}$$

and hence no such polynomial $f$ can pass through a point like $\left( - \frac{1}{2}, \frac{1}{3} \right)$.