Darrel C Jones analysis

I.) 8 x x R N x x x R
    7 x P P P B x P x
    6 P R B x Q x x x
    5 P x P x x x x B
    4 x B x P x N R x
    3 x N x P P x P P
    2 P x P K x N K x
    1 Q P x P x x P x
    - a b c d e f g h

(The diagram is labelled to make identification of the squares easier.)

The only piece that can attack a Q without the Q attacking it is an N, so
the queens are both under attack by N's. The only possible squares for
this are a1 and e6, with one N necessarily on b3. Thus b8 and g8 are
empty. Since a Q attacks b2, no K can be there (or it would attack the
Q), so b2 is empty. P's can only attack R's, N's, and other P's without
being attacked back, and we're given they're moving "up the board", so
e8, a7, f7, h7, d5, f5, e4, a3, c3, and h1 are empty. (All of these
either attack only B's, Q's, K's, and empty squares, except for d5, f5,
and e4. The Q on e6 means d5 and f5 are empty, and this requires e4 be
empty too.) Now a K on f8 cannot attack any square not being attacked by
it, so f8 is empty. The K's must be on d2 and g2. Now f6, c3, f3, h2, e1,
and f1 are all empty, as they would otherwise contain pieces being
mutually attacked by pieces known to be present. The only possible
squares for B's are e7, c6, d6, h5, and b4. B's on e7 and d6 would
mutually attack one another, so they cannot both be real. Since there are
4 B's altogether, c6, h5, and b4 are occupied. And if d6 were occupied,
either the P on c5 would be present and mutually attack d6, or it would
be absent and d6 would mutually attack d4. Thus d6 is absent and e7 is
present. Similar reasoning shows the P on e5 is present to prevent b4
from mutually attacking e7. Now the P on b5 cannot be there, as it would
mutually attack the real piece on c6. The P's on e5, a4, and c4 cannot
attack anything, so those squares are also empty. 15 P's must be placed,
and only 16 squares remain as possibilities (b7, c7, d7, g7, a6, a5, d4,
h4, e3, g3, h3, a2, c2, b1, d1, and g1), so no more than one of those 16
can be empty. Looking at the R's, there are 6 possible squares for the 4
of them, and R's on both a8 and c8, or on both h8 and f8, would mutually
attack each other, so the R's on b6 and g4 are real. If the R on c8
weren't there, than the P on d7 couldn't be there (as it could attack
nothing), and the P on c7 couldn't be there (as nothing could attack it).
But at least one of those P's is present, so the R on c3 must be, too,
with a8 empty. The only pieces an R on f8 could attack would attack it,
so f8 is empty, and an R sits on h8. Since it and the R on c8 do not
attack each other, an N is on d8. If the P on g7 weren't there, then a P
must be on d4, so the only piece that could attack the R on h8 would be
an N on g6. It would also require the P's on h4 and g7 be real, so N's
would have to be on g5 and  and f2 (so that those P's could attack
something). But this would give a total of five N's, too many. Therefore
a P really does stand on g7. The only piece that can attack this P is the
Q on a1 or the R on g4. If it were the Q on a1, the P on d4 wouldn't be
present. So the other P's would, and this requires the N's on g5 and f2
to be the last pieces on the board (to give the P's on h4 and g1
something to attack.) But now the P on e3 has nothing to attack, so the P
on g7 must be attacked by the R on g4. So the N's on g5 and g6 can't be
on the board, giving the P on h4 nothing to attack. So the other
remaining P's, including the ones on g3 and g1, are there, requiring the
remaining N's to stand on f4 and f2 to give them something to attack.