Pairs of integer pairs with same lcm, gcd and mean The problem is to find all pairs of two distinct pairs(up to permutation) of integer(!) numbers $(a, b)$ and $(c, d)$ s.t. $$\operatorname{lcm}(a, b) = \operatorname{lcm}(c, d)$$
$$\gcd(a, b) = \gcd(c, d)$$ and 
$$\frac{a + b}{2} = \frac{c + d}{2}$$

It is easy to show that if both LCM and GCD are equal, then two pairs have the same product in absolute value and the same sum AND the same GCD.

There was a similar question on MathSE about natural numbers, answer to which is that such distinct pairs dont exist: Prove two distinct pairs of natural numbers with these properties do not exist

One such pair is ((-6; 35), (14;15)): their sums and LCMs and GCDs are all equal pairwise. How to find all such pairs?
Given $\operatorname{lcm}(x,y)\cdot\operatorname{gcd}(x,y)=|xy|$, we have $\operatorname{lcm}(a,b)\cdot\operatorname{gcd}(a,b)=\operatorname{lcm}(c,d)\cdot\operatorname{gcd}(c,d)$, so $|ab|=|cd|$. Given $\frac{a+b}{2}=\frac{c+d}{2}$, we have $c=(a+b)-d$. When substituted into $|ab|=|cd|$, this gives the two quadratics $d^2-(a+b)d\pm ab=0$. These have the two pairs of solutions $d=a,b$ (when $c,d$ are trivially a permutation of $a,b$) and $d=\frac{(a+b)\pm\sqrt{\Delta}}{2}$, where $\Delta$ is the discriminant $a^2+6ab+b^2$. 

In the nontrivial case, $d$ is integral if $(a+b)\pm\sqrt{\Delta}$ is integral (and even), therefore $\Delta$ is a square. Hence, for some integer $k$, the triple $(a,b,k)$ is a solution of the 3-variable, 2nd-degree Diophantine equation $a^2+6ab+b^2=k^2$. This can be solved with a method analogous to finding Pythagorean triples: consider the intersections of the hyperbola $a^2+6ab+b^2=1$ with the line $a=m(b-1)$. The first intersection is $(1,0)$ and the second intersection is guaranteed to be rational when $m$ is integral, which paves the way for integer solutions when clearing denominators. The second solution is $b=\frac{(m+1)(m-1)}{m^2+6m+1},a=\frac{-2m(3m+1)}{m^2+6m+1}$, which we can substitute into $a^2+6ab+b^2=1$ and multiply through by $(m^2+6m+1)^2$, to give solutions for $(a,b,k)$, and by extension $(a,b,c,d)$ when substituted into the earlier equations. So the set of nontrivial solutions up to multiples and permutations is, for $m\in\mathbb{Z}$

$a=-2m(3m+1)\\
 b=(m+1)(m-1)\\
c=-(3m+1)(m+1)\\
d=2m(1-m)$

For example, $m=-5$ gives the solution $(-140,24,-56,-60)$, which is a multiple and permutation of the solution $(-6,35,14,15)$ mentioned by the asker. This encompasses all solutions.
Given $\gcd(a,b)=1$, since $ab=cd$, assume $$(-a,c)=x, (-a,d)=y, (b,c)=z, (b,d)=w$$
we have $$a=-xy, b=zw, c=xz,d=yw$$ so that $zw-xy=xz+yw$ or $(z-y)w=x(y+z)$
so
$x\mid z-y$ and $w\mid y+z$;
$z(w-x)=y(x+w)$
$z\mid x+w$ and
$y\mid w-x$.

Now let's assume that $w$ is largest (x,y,z,w are all positive integers) among x,y,z,w
since $w\mid y+z$, we must have $w=y+z$
so next we got $z=x+y$
So given any integer number pairs $(x,y)$, we have $$x=x,y=y,z=x+y,w=2y+x$$
$$a=-xy, b=(x+y)(x+2y), c=x(x+y), d=y(x+2y)$$ is a result.
And given any another integer $f$, we have
$$a=-xyf, b=(x+y)(x+2y)f, c=x(x+y)f, d=y(x+2y)f$$ is a result