For which integers $a,b,c,d$ does $\frac{a}{b}+\frac{c}{d} = \frac{a+c}{b+d}$? 
For which $a,b,c,d \in \mathbb{Z}$ does $\frac{a}{b}+\frac{c}{d} = \frac{a+c}{b+d}$?

This is actually the question I meant to ask in a previous question that I asked here.

What about $a,b,c,d \in \mathbb{R}?$

 A: If you clear the fractions you get $abd+ad^2+b^2c+bcd=abd+bcd$ or $$ad^2+b^2c=0$$  As long as none of the denominators are zero, you are set.
A: Multiply both sides by $bd(b+d)$ and you get:
$$\begin{align}ad(d+b) + cb (d+b) = bd(a+c)&\iff\\
ad^2 + abd + cbd + cb^2 = b d(a+c)&\iff\\
 ad^2 +cb^2 = 0&\iff\\
 ad^2=-cb^2
\end{align}$$
Given $b,c$, you need $d^2$ to be a factor of $cb^2$ and then $a=-\frac{cb^2}{d^2}$.
Given non-zero integers $m,n,p,q$, with  you have:
$$(a,b,c,d)=(pm^2,mq,-pn^2,nq)$$
This gives all solutions, excluding the case where $m=-n$.
The only solution with $\gcd(a,b)=1$ and $\gcd(c,d)=1$ are the cases where $m,n=\pm 1$, and then you are dealing with the solutions:
$$\frac{a}{b}+\frac{-a}{b}=\frac{a+(-a)}{b+b}$$
A: I am adapting my answer to a previous question here.
$$\frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}$$
We will keep careful track of which quantities may be zero. To start, we know that $b, d, b+d \neq 0$ because they are in the denominator. 
Let's multiply both sides by  $b\cdot d \cdot (b+d)$:
$$\begin{align*}ad(b+d) + bc(b+d) &= (a + c)bd\\
abd + ad^2 + b^2c + bcd &= abd + bcd\\
abd + ad^2 + b^2c + bcd - abd - bcd &= 0\\
\fbox{$ad^2 + b^2 c$ = $0$}
\end{align*}$$
Here's an example assignment that satisfies this equation: 
$$\begin{align*}a &= 5\\b&=3\\c&=-20\\d&=6\\\end{align*} $$

We can parameterize this set of solutions with three parameters, $\alpha, \beta_1, \beta_2 \in \mathbb{R}$ and three minimal constraints $\beta_1 \neq 0, \beta_2 \neq 0$, and $\beta_1 \neq \beta_2$. For any such triple, a solution is given by 
$$\begin{align*}
a &\equiv \alpha\\
b &\equiv \frac{1}{\beta_1}\\
c &\equiv -\alpha\frac{\beta_1^2}{\beta_2^2}\\
d &\equiv -\frac{1}{\beta_2}\\
\end{align*}$$

Conversely, given any solution $\langle a, b, c, d\rangle$ with $\frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}$, we can solve uniquely for parameters $\langle \alpha, \beta_1, \beta_2\rangle$ meeting these constraints.

For fun, here is a nomogram that allows you to find such collections of numbers:

You can draw any line passing through, say, the B and C axes, and any line passing through the A and D axes. If the two lines meet at the same point on the central vertical axis, then the corresponding points $(a, b, c, d)$ satisfy the equation.
In the picture, the dashed lines show an example. Here, we see that the equation is satisfied when $a = 4$, $b=3$, $c=-1$, and $d=1.5$.
Also, if you specify any four of the following five variables—a value for $a$, a value  for $b$, a value for $c$, a value for $d$, a spot on the central axis— you can "solve" for the remaining point.
