Solution:
The polynomial $x+2004$ certainly meets the two conditions. In fact, this is the only one. We will prove this using three ingredients: the infinitude of primes, the Rational Roots Theorem for polynomials, and the approximation principle that $x^{n}$ dominates any polynomial of lower degree, for large enough $x$.

Note that the only monic constant polynomial is $f(x)=1$, which fails Condition 1; and that the only monic degree 1 polynomial satisfying Condition 1 is $f(x)=x+2004$. Thus, we need to eliminate all polynomials of degree 2 or more. To this end, it is sufficient to show that, given any monic polynomial $f(x)$ with integer coefficients of degree 2 or more, there exists an integer $a$ such that $f(x)+a=0$ has an irrational solution $x$ (for then $f(x)=-a$ is rational for an irrational number $x$).

Let $f(x)=x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0}$ be a polynomial with integer coefficients, with $n \geq 2$. It may be the case that $f(x)$ has no real roots, for example, if $n$ is even and the graph of $y=f(x)$ lies above the $x$-axis. But certainly, if $a$ is a sufficiently large negative integer, we can guarantee that $f(x)+a=0$ will have at least one real solution. In fact, by further making $a$ a larger negative number, we can ensure that, say, the largest of the solutions of $f(x)+a=0$ has absolute value bigger than 1: $|x|>1$.

Moreover, regardless of how large a negative number $a$ needs to be, we can choose $a$ so that $c_{0}+a=-p$ where $p$ is prime. This is because there are infinitely many prime numbers. Now we can apply the Rational Roots Theorem, according to which all rational solutions $\frac{r}{s}$ of the monic integer coefficient polynomial $f(x)+a$ must satisfy: $s$ divides the leading coefficient of $f(x)$ and $r$ divides the last (free term) of $f(x)$; in other words, $s$ divides 1 and $r$ divides $p$. Since $p$ is prime, this gives only four possible rational solutions: $x= \pm 1, \pm p$. Since we have ensured that $|x|>1$, we are left with $x= \pm p$.

Let $g(x)=f(x)+a$. From the well known inequalities of absolute values $|y+z| \geq |y|-|z|$ and $|y+z| \leq |y|+|z|$, we obtain:
$$|g(x)|=\left|x^{n}+c_{n-1} x^{n-1}+\ldots+c_{1} x-p\right| \geq \left|x^{n}\right|-\left|c_{n-1} x^{n-1}+\ldots+c_{1} x-p\right|$$
and as long as $|x|>1$ and $n \geq 2$:
$$\begin{aligned} \left|c_{n-1} x^{n-1}+\cdots+c_{1} x-p\right| & \leq \left|c_{n-1}\right||x|^{n-1}+\cdots+\left|c_{1}\right||x|+p \\ & \leq \left(\left|c_{n-1}\right|+\cdots+\left|c_{1}\right|\right)|x|^{n-1}+p^{n-1} \end{aligned}$$
If we let $S=\left|c_{n-1}\right|+\cdots+\left|c_{1}\right|$, we can put everything together:
$$|g( \pm p)| \geq p^{n}-(S+1) p^{n-1}=p^{n-1}(p-(S+1))$$
Since $S$ is fixed, we can choose the prime $p$ large enough so that $p>S+1$, and hence the quantity $p-(S+1)$ is positive. Therefore, $g( \pm p) \neq 0$.

Thus $g(x)$ has a real zero $x$, which cannot be rational since the only possibilities for rational zeros $\pm p$ fail to be zeros by the above. We conclude that $x$ is an irrational root of $g(x)$, whereas $f(x)=-a$ is an integer, hence rational. This contradicts Condition 2, and eliminates all polynomials of degree 2 or more.

Finally, we are left with only one possible solution: $f(x)=x+2004$.