Let $A = \frac{a}{a+1} + \frac{a}{a+2} + \frac{a}{a+3} + \frac{a}{a+4}$. Clearly, we have $0 < A < 4$.

Solution using GCD: Write $a = \frac{p}{q}$ with $p, q$ relatively prime positive integers. Then can write
$$A = \frac{p}{p+q} + \frac{p}{p+2q} + \frac{p}{p+3q} + \frac{p}{p+4q} = \frac{2p(2p+5q)(p^2+5pq+5q^2)}{(p+q)(p+2q)(p+3q)(p+4q)}$$
Now note that $p^2 + 5pq + 5q^2$ is relatively prime to $p+q, p+2q, p+3q$ and $p+4q$, since
$$\begin{align*} (p+q, p^2+5pq+5q^2) &= (p+q, 4pq+5q^2) = (p+q, 4p+5q) = 1 \\ (p+2q, p^2+5pq+5q^2) &= (p+2q, 3pq+5q^2) = (p+2q, 3p+5q) = 1 \\ (p+3q, p^2+5pq+5q^2) &= (p+3q, 2pq+5q^2) = (p+3q, 2p+5q) = 1 \\ (p+4q, p^2+5pq+5q^2) &= (p+4q, pq+5q^2) = (p+4q, p+5q) = 1 \end{align*}$$
It follows that
$$\frac{2p(2p+5q)}{(p+q)(p+2q)(p+3q)(p+4q)} = \frac{A}{(p^2+5pq+5q^2)}$$
is an integer. This is a contradiction since $0 < A < 4$ and $p^2 + 5pq + 5q^2 > 4$.

Solution using polynomials: Let
$$\begin{align*} f_A(x) &= (x+1)(x+2)(x+3)(x+4) \left( \frac{x}{x+1} + \frac{x}{x+2} + \frac{x}{x+3} + \frac{x}{x+4} - A \right) \\ &= (4-A)x^4 + 30(3-A)x^3 + 35(2-A)x^2 + 50(1-A)x - 24A. \end{align*}$$
Then, for $A = 1, 2$ and $3$, we have
$$\begin{align*} f_1(x) &= 3x^4 + 60x^3 + 35x^2 - 24, \\ f_2(x) &= 2(x^4 + 15x^3 - 25x - 24), \\ f_3(x) &= x^4 - 35x^2 - 100x - 72. \end{align*}$$
It is easy to show that these polynomials do not have rational roots using the rational root lemma.