Solution:

Let us rearrange the first equation to read
$$a+b=-(c+d)$$
and the second to give
$$\begin{aligned} \frac{1}{a}+\frac{1}{b} & =-\frac{1}{c}-\frac{1}{d} \\ \frac{a+b}{a b} & =-\frac{c+d}{c d} \end{aligned}$$
If $a+b \neq 0$, then we can divide (1) by (2) to yield $a b=c d$. So we conclude that
$$a+b=0 \quad \text{ or } \quad a b=c d$$
In exactly the same manner, we derive that
$$a+c=0 \quad \text{ or } \quad a c=b d$$
and
$$a+d=0 \quad \text{ or } \quad a d=b c \quad \text{ (this equation will not be needed). }$$
For the sake of contradiction, we are assuming that $a+d \neq 0$. Suppose that $a+c=0$ holds; since $a+b+c+d=0$, we get $b+d=0$, whereas we know $a \leq b$ and $c \leq d$. We conclude that $a=b$ and $c=d$, whence $a+d=a+c=0$. Thus $a+c \neq 0$, and for similar reasons $a+b \neq 0$.
Multiplying together the equations $a b=c d$ and $a c=b d$ and canceling the nonzero (why?) factor $b c$ gives $a^{2}=d^{2}$. So either $a=-d$, which is what we assumed impossible, or $a=d$, which yields the absurdity $a=b=c=d=0$.

Solution:

Consider the polynomial
$$f(x)=(x-a)(x-b)(x-c)(x-d)$$
with roots $a, b, c, d$. The cubic coefficient of $f$ is $-(a+b+c+d)=0$; the linear coefficient is
$$-(a b c+a b d+a c d+b c d)=-a b c d\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=0 .$$
Consequently $f$ has the form $f(x)=x^{4}+p x^{2}+q$, and there are complex numbers $u$ and $v$ such that
$$f(x)=\left(x^{2}\right)^{2}+p x^{2}+q=\left(x^{2}-u\right)\left(x^{2}-v\right)=(x+\sqrt{u})(x-\sqrt{u})(x-\sqrt{v})(x+\sqrt{v}) .$$
Since $f$ has only real roots, $u$ and $v$ must be real and nonnegative. WLOG $u \leq v$; then $-\sqrt{v} \leq-\sqrt{u} \leq \sqrt{u} \leq \sqrt{v}$ and we can identify the two factorizations of $f$ :
$$a=-\sqrt{v}, \quad b=-\sqrt{u}, \quad c=\sqrt{u}, \quad d=\sqrt{v} .$$
Hence $a+d=0$.