In district I, A wins (in most Condorcet methods, anyhow).
In district II, (same as district I but the roles of B and D are swapped),
A also wins.
But in the combined 2-district country,
C is the Condorcet winner (beats A by 8:6, beats B and D by 9:5).

Consequently,
some Condorcet methods cannot be "counted in Precincts."
For example, the Smith,IRV method
invented by Woodall
is Condorcet but I see no
efficient way to count it in precincts if there are a large number
of candidates (say 100 candidates, if you want something concrete to
think about – see any feasible way to count it other
than a centralized count? I don't).

However, the Condorcet methods that depend only on the matrix of "pairwise totals"
can be counted in precincts (despite the nonadditivity paradox illustrated above!)
in the sense that each precinct can find and publish its pairwise matrix
and the summed matrix
will be used for the whole country.

Here is a second example now with only 3 candidates:

District I

#voters

Their Vote

30

A>B>C

20

B>C>A

20

C>A>B

20

B>A>C

District II

#voters

Their Vote

21

A>C>B

20

C>B>A

20

B>A>C

A wins in district I with every Condorcet method since
A beats B pairwise 50:40 and also beats C pairwise 50:40.

In district II, the situation is completely symmetric between A, B, and C
except for a single extra ballot of type "A>C>B." Therefore, A wins
in every Condorcet method ever seriously proposed, although
A is not a "beats-all winner" because there is a cycle.

But in the combined country, B wins
with every Condorcet method since
B beats A pairwise 80:71 and also beats C pairwise 90:61.

For those who cannot get enough, here is a third example again with only 3 candidates:

District I

#voters

Their Vote

3

A>B>C

4

B>C>A

3

C>A>B

District II

#voters

Their Vote

2

A>B>C

3

B>A>C

In district I, the situation is completely symmetric between A, B, and C
except for a single extra ballot of type "B>C>A." Therefore, B wins
in every Condorcet method ever seriously proposed, although
B is not a "beats-all winner" because there is a cycle.

In district II, B is ranked top by 60% of the voters hence
wins in
in every Condorcet method.
(Also wins with Borda and virtually every other voting method.)

But in the combined country, A wins
with every Condorcet method since
A beats B pairwise 8:7 and also beats C pairwise 8:7.