Communication between observers in gravity wells Suppose two friends live 100 light years apart. Can they use gravitational time dilation to reduce the 200-year roundtrip lag they experience during their phone calls?
Specifically, how would the lag (from their perspective) change if they move their houses close to the event horizons of their individual black holes? What if they instead drag a huge black hole (with Schwarzschild radius of ~50 ly) to the midpoint between them, so that they both end up near the event horizon of the same black hole?
 A: 
Suppose two friends live 100 light years apart. Can they use gravitational time dilation to reduce the 200-year roundtrip lag they experience during their phone calls?

The OP describes two different scenarios, with very different spacetime geometries, so I'll address them separately.

 First scenario: individual black holes 
Let $D$ denote the distance between the two friends, with $D\sim 100$ light years. Suppose that each friend is near a black hole with Schwarzschild radius $R\ll D$, as shown here:

I'll show the analysis in some detail, because the answer is not obvious (at least not to me). The key is to think about the relationship between each friend's proper time and the coordinate time. Since the two black holes are far away from each other, we can use the same relationship that we would use for an isolated Schwarzschild black hole:
$$
 \Delta \tau = \sqrt{1-\frac{R}{r_0}}\,\Delta t,
\tag{1}
$$
where $\Delta \tau$ is the elapsed proper time for an friend hovering at a fixed coordinate-radius $r_0>R$ above the black hole, and $\Delta t$ is the usual coordinate-time. Equation (1) says that for a given elapsed coordinate-time $\Delta t$, the elapsed proper time $\Delta\tau$ experienced by the hovering friend can be made as small as we like, just by putting the friend very close to the horizon $(r_0\gtrsim R)$. The problem of how the friend might maintain that hovering position (e.g., with a rocket) will not be addressed here.
That doesn't answer the question yet, because we still need to estimate the elapsed coordinate-time $\Delta t$ for a signal to make a round trip. Since $R\ll D$, we can analyze this using the metric for an isolated black hole when the signal is near either hole, and using the metric of flat spacetime in the vast space between the two holes. Since the metric of an isolated black hole approaches flat spacetime far from the hole, we can do this instead: we can use the isolated-black-hole metric for whichever black hole is nearest to the propagating signal at any given time. Assuming that the black holes have the same mass, this makes the problem nicely symmetric: we can just calculate the elapsed time for the signal to reach the halfway point, and then multiply by $4$. (One factor of $2$ converts halfway to a full one-way trip, and the other factor of $2$ converts the one-way trip to a two-way trip.) Near one of the black holes, the coordinate-speed of a lightlike radial worldline is (in units where the limiting speed is $1$)
$$
 \frac{dr}{dt} = 1-\frac{R}{r}
\tag{2}
$$
where $r$ is the radial coordinate. (The friend is hovering at $r=r_0$.) For $r>R$, equation (2) is solved by
$$
 t = r + R\ln(r-R) + \text{constant},
\tag{3}
$$
so the elapsed coordinate-time $\Delta t$ required for the signal to go from a point $r_0$ near the hole (where the friend is hovering) to the halfway point $r=D/2$ is
$$
 \Delta t 
 = 4\Delta r + 4R\ln\left(\frac{D/2-R}{r_0-R}\right)
 = 4\Delta r + 4R\ln\left(\frac{\Delta r}{r_0}+A\right) - 4R\ln A
\tag{4}
$$
with 
$$
\Delta r := \frac{D}{2} - r_0
\hskip2cm
A := 1-\frac{R}{r_0}.
\tag{5}
$$
Combine this with (1) to conclude that the round-trip time experienced by the hovering friend is
$$
 \Delta \tau = 4\left(\Delta r + R\ln\left(\frac{\Delta r}{r_0}+A\right) - R\ln A\right)
  \sqrt{A}
\tag{6}
$$
Since $(\ln A)\sqrt{A}$ goes to zero as $A\to 0$, this reduces to
$$
 \Delta \tau \approx 4\left(\Delta r + R\ln\frac{\Delta r}{r_0}\right)
  \sqrt{A}
\tag{7}
$$
for sufficiently small $A$. Since we can make $A$ as close to zero as we like by putting the friend's radial coordinate $r_0$ closer and closer to $R$ (that is, closer and closer to the event horizon of the black hole), we can make the signal's round-trip time experienced by that friend as small as we like. 
Conclusion: In this scenario, the two friends can communicate with arbitrarily small lag, by hovering sufficiently close to the horizons of their respective black holes.

 Second scenario: single enormous black hole 
In this scenario, we have a single black hole with the enormous Schwarzschild radius $R\sim D/2$, with the two friends hovering on opposite sides of the black hole, as shown here:

They can communicate with each other by sending signals (again assumed to be lightlike) around the black hole, and if the hovering radius $r_0$ is just barely greater than $3R/2$, then the signal's orbit will be nearly circular. Equation (1) still applies here, but instead of equations (2)-(4), we can use the coordinate-time orbital period for a lightlike circular orbit with radius $3R/2$, namely
$$
 \Delta t \approx \frac{3\pi R}{\sqrt{A}}
\hskip2cm
 A := 1-\frac{R}{r_0} \approx 1-\frac{R}{3R/2}.
\tag{8}
$$
One period of a circular orbit is the same as one round-trip time between friends on opposite sides, so combining equations (1) and (8) shows that the round-trip time for friends hovering near $r_0\approx 3R/2$ is
$$
 \Delta \tau \approx \sqrt{A}\,\frac{3\pi R}{\sqrt{A}} \approx \frac{3\pi}{2} D.
\tag{9}
$$
For two friends separated by a distance $D$ in flat spacetime, the round-trip time would be $2D$ (again units where the fundamental speed limit is $1$). Since $3\pi/2 > 2$, the round-trip time is actually longer in this single-black-hole scenario than it would be in flat spacetime. We can't fix this by trying to send the signal inward, short-cutting the circular orbit, because then the signal would spiral into the black hole without ever reaching the other hovering friend. That's because $r=3R/2$ is the radius of the innermost circular orbit around a (non-rotating) black hole. Light can escape from within that radius, as long as $r>R$, but only if it is directed away from the black hole. Granted, I haven't considered the possibility that the time could be reduced by using such outward-directed signals, but that seems intuitively unlikely. 
Conclusion: In this scenario, the two friends (probably) cannot reduce the lag in their communication, compared to what it would be with no black holes at all.

In summary, For minimizing the communication lag experienced by the two friends, two small black holes is better than none, and one big black hole is probably worse than none.
