The Delden version from the Jaspers original question asks for solutions to this kind of fractions,
1/Π(pi) = Σ (si/pi), where pi are distinct primes and si=+1 or -1, only.

Example: 1/(3*5*7*53*107)=1/3-1/5-1/7+1/53-1/107, n=5

Jan van Delden claims that:

For n=2 the sign pattern (+1,-1) will give only
solutions for p[1]=2

For n=3 the sign pattern (+1,-1,-1) and (-1,+1,+1) will give solutions
for p[1]=2; the latter is the only one for p[1]>2

For n=4 the sign pattern (+1,-1,-1,+1), (-1,+1,+1,-1), (-1,+1,+1,+1),
(+1,-1,-1,-1) will give only solutions for p[1]=2

Some of my results.

I)

There are no solutions with p[1]>2 for n
even.

If n even the sign pattern can be split
up in 2-terms of the form (+1,+1), (+1,-1), (-1,-1), or (-1,+1).

After forming a combined fraction for
each of these 2-terms, the numerator will be even. Combining all
these 2-terms will therefore have an even numerator, which can't be
1.

II)

The only 2-term is [2, -3] (where I
included the sign in the solution).

III)

If n=3 the number of solutions with
fixed p[1]=p are finite, in fact if p>2 the upperbound is
pi(2p)-pi(p) and if p=2 there are 2 solutions.

If p[1]=2 then we have the only
solutions [2,-3,-7] and [-2,3,5].

A 3-term with sign pattern (-1,+1,+1)
with (p,q,r) as denominator with p<q<r will have r=(pq-1)/(q-p).
Solving r>q gives q<p+sqrt(p^2-1)<2p. So (p,q) defines r and r>q
fixes the number of q allowed. Since p<q<2p the upperbound is as
given if p>2. If p=2 the only possible solution is q=3, this gives
r=5. (The sign pattern (+1,-1,-1), only solutions for p=2, gives
q=3, r=7 in a similar manner).

Furthermore r/p is uniform decreasing as
a function of q, so the maximum for r is attained for the smallest q
possible, so for p=2 we get r/p=2.5. If p>2 then we are looking for
twin primes to achieve a maximum for r (resulting in
r=p+(p-1)(p+1)/2), hence the maximum value of r will be O(p^2/2).

IV)

There's a simple way to eliminate a lot
of sign patterns from the 2^n possible patterns.

Assume that the p[i] are in ascending
order: p[i]<p[j] if i<j.

The patterns (+1,+1) and (+1,-1) will
result in a + contribution to the total sum.

The patterns (-1,-1) and (-1,+1) will
result in a - contribution to the total sum.

[In fact the signs used can be further
apart then 1 position]

Assume our total sum is split in two
parts (and we have n>2), the first part being a/b the second c/d
after combining the unit fractions.

Then we have ad+bc=1, since b and d are
(both) a product of primes these terms are positive. We get
necessarily that sign(a) and sign(b) are opposite.

[If they are both positive then ad+bc>1,
both negative then ad+bc<0].

The above rule should hold up using all
possible splits for any pattern.

So for instance (n=3):

(+1,+1,+1) can't be split in a form with
a sign change, therefore not allowed

(+1,+1,-1) can be split as (+1,+1)(-1)
giving (+1,-1) a sign change. But also (+1)(+1,-1) giving (+1,+1)
no sign change, so not allowed!

This can ofcourse be
seen differently by combining the first +1 with the last -1 giving
(+1,+1) directly.

(+1,-1,+1) Not allowed, same reason

(+1,-1,-1) both splits result in
(+1,-1) hence allowed.

The other patterns start with -1, use
symmetry, so also (-1,+1,+1) allowed.

[You may verify that (-1,+1,+1) is the
only pattern with p[1]>2]

Of the above sign-patterns we also see
that (+1,+1,+1) ,(+1,+1,-1) and (+1,-1,+1) give a + contribution to
the total sum. The opposite patterns give a - contribution and are
therefore also not allowed. This contribution can be used if testing
for splits for n>3 of size 3. The splits for n=3 that are admissible
for n=3 are inconclusive if used as patterns in n>3!