Is there always a frame in which spatially separated events are simultaneous? In relativity, if two events are simultaneous in a specified frame, they cannot be simultaneous in any other frame. 
My question is this: given any two events, is there always a frame in which these two events are simultaneous? For example, if I drop a blue ball on one side of a tennis court, and my friend drops a red ball on the opposite side of the court one day later -- from my frame one day later -- is there a frame in which the blue and red balls hit the ground simultaneously? 
 A: Joshua's answer is correct, and I see you have accepted it, but let me try and show how this emerges from the Lorentz transformations.
Let's choose our coordinates so that the earlier point is at the origin, then the second point is at $(t, x)$. That is the two events are separated by a time $t$ and a distance $x$. Now let's try and find a frame where the two points are simultaneous. The Lorentz transformations tell us:
$$\begin{align}
t' &= \gamma (t - \frac{vx}{c^2}) \\
x' &= \gamma (x - vt)
\end{align}$$
The first point $(0, 0)$ just transforms to $(0, 0)$ so the condition for the events to be simultaneous is that $t' = 0$, so we have:
$$ t' = \gamma (t - \frac{vx}{c^2}) = 0 $$
and this is true only if:
$$ t = \frac{vx}{c^2} $$
or:
$$ v = c^2\frac{t}{x} $$
But the maximum velocity $v$ can have is $c$, so for the frame to exist we have the inequality:
$$ c^2\frac{t}{x} < c $$
or with a quick rearrangement:
$$ \frac{x}{t} > c $$
And here is our result. The ratio $x/t$ has the dimensions of a velocity, and this velocity must be greater than the speed of light. In other words a frame where the two events are simultaneous can only exist if the two events are spacelike separated.
A: 
Is there always a frame in which spatially separated events are
  simultaneous?

The answer is no.
Two events that are spatially separated in one frame of reference
(1) will be co-located in another frame of reference and not simultaneous in any frame if the interval is time-like
(2) will be simultaneous in another frame of reference and not co-located in any frame if the interval is space-like .
(3) will be neither co-located nor simultaneous in any other frame if the interval is light-like.
Time-like interval
If the interval is time-like, the separation in time, $|c\Delta t|$, is larger than the separation in space $|\Delta x|$:
$$|c\Delta t| \gt |\Delta x|$$
Thus, there is a frame of reference in which $\Delta x' = 0$; the two events are co-located in this frame.
Space-like interval
If the interval is space-like, the separation in time is less than the separation in space:
$$|c\Delta t| \lt |\Delta x|$$
Thus, there is a frame of reference in which $c\Delta t' = 0$; the two events are simultaneous in this frame.
Light-like interval
If the interval is light-like the separation in time equals the separation in space:
$$|c\Delta t| = |\Delta x|$$
Thus, in all frames of reference, the events are neither co-located nor simultaneous, i.e.,
$$|c\Delta t'| = |\Delta x'|$$
All of this follows directly from the Lorentz transformation.  Let's take your example of two events with spatial separation of a tennis court so
$$|\Delta x| = 78\mathrm m$$
Light travels this distance in $\Delta t_c = \frac{78}{300 \cdot 10^6} = 260\mathrm{ns}$
Thus, if the two events occur within 260ns in this frame of reference, the events have space-like interval and are thus simultaneous in another, relatively moving reference frame of reference.
Since, in your example, the events occur 1 day apart, the events have time-like interval and cannot be simultaneous in any reference frame.
A: We must look at the spacetime interval between the two events.  Two events separated by a distance magnitude $\Delta r$ and a time interval $\Delta t$ in a frame are said to be space-like if $c^2\Delta t^2 < \Delta r^2$ (i.e., the distance between the events is too great for light emitted by one event to be seen by the other) and are said to be time-like if $c^2\Delta t^2 > \Delta r^2$ (i.e., the distance between the events is small enough that light emitted by one event can be seen by the other).  Events with $c^2\Delta t^2 = \Delta r^2$ are called light-like.  These events are separated by just the right amount of spatial distance that light from one event will reach the other right as it is happening.
Space-like events cannot have a causal relationship (i.e., one event is influence by the other) because the spatial separation between the two is too great for information from one of the events to reach the other before it happens.  For these events, there is a reference frame where the two events happen simultaneously but no reference frame where the two events happen at the same location.
Time-like events can have a causal relationship since the spatial separation is small enough that light/information can travel from one event to the other.  For such events there exists a reference frame where the two events happen at the same location but no reference frame where they happen at the same time.
So, in short, if two events are separated by a space-like interval there does exist a frame where they happen simultaneously.  If two events are separated by a time-like interval there does not exist a frame where they happen simultaneously.  In your tennis ball example, since the spatial separation is much smaller than the time separation (i.e., enough time elapses between the events that light from one event can be viewed by the other) we have a time-like interval and the events cannot be viewed simultaneously in any reference frame.  At least, that is assuming you're talking about a standard tennis court.  If you had a tennis court that was longer than one light-day then you could have a space-like interval between the two events.
A: No. Causally related events can never be viewed as simultaneous in any frame.
A: This is only possible if both events are independent of each other and satisfy the condition...
$∆t/∆x=v/c^2$
$∆t$ implies time interval between events,
$∆x$ implies difference in coordinates of events.
