$5+\sqrt{3}$

Call the sum of the six distances $S$. First, in order to achieve the desired maximum, there MUST be one distance greater than one; otherwise the best possible configuration is simply the four points being the four vertices of a regular tetrahedron. Say $AB$ is the largest distance of the six; then $1<AB<2$, the latter inequality because of the triangle inequality; if $AB = 2$, then $C$ and $D$ would have to be the midpoint of $AB$, and then $S$ will only be $6$. Draw 2 spheres each of radius $1$ with centers at $A$ and $B$. Observe that points $C$ and $D$ must lie in the interior of both spheres, because their distances from $A$ and $B$ cannot be greater than $1$. Observe further that in order to maximize $S$, $C$ and $D$ must actually be on the circle that is the intersection of the spheres, and must be opposite each other on this circle, forming a diameter. This will make $AC$, $AD$, $BC$, and $BD$, all equal to $1$, and will also maximize $CD$ within the given space. Now we know that the four points are coplanar and form a rhombus of side length 1, and the only remaining unknown is the length of $AB$ itself. We are trying to make $AB+CD$ as large as possible, so we notice that they are diagonals our rhombus, and that $\frac{AB}{2}$ and $\frac{CD}{2}$ are the legs of a right triangle with hypotenuse $1$. Sum of the lengths of legs of a right triangle with fixed length is maximized when the triangle is isosceles, but our triangle can't be isosceles, as that would make $CD>1$. Then we want the triangle as close to isosceles as possible, which happens when $CD=1$. Wait- so $\triangle ACD$ and $\triangle BCD$ are equilateral! We have two equilateral triangles of side $1$ which share a side to form a rhombus, and so $AB$ is just two times the height of one of these, or $\sqrt{3}$. This brings us to our final answer of $S = \boxed{5+\sqrt{3}}$, since all of the other lengths are $1$.