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Here's the start of a proof.

We will say that two arcs overlap twice if their union covers the entire circle. First, note that you can assume that there are no two red arcs that overlap twice.

Proof: Suppose that there were two blue arcs that overlap twice and two red arcs that overlap twice. Then, we could delete all four of these arcs and find a point covered by $n-2$ of the remaining arcs. Two of the deleted arcs also cover this point, and we're done. By symmetry, we can assume that there are no two red arcs that overlap twice.

Now, note that we can assume that no red arc is contained in another. If there were, we could shrink the larger one so that they overlap, but neither is contained in the other.

These two observations mean that we can order the red arcs clockwise around the circle. Let the red arcs be $(L_i, R_i)$, with $L_i$ being the left (anticlockwise) endpoints and $R_i$ be the right (clockwise) endpoints. Then $L_i$ and $R_i$ are also ordered clockwise around the circle.

Now, we can assume that the only blue arcs we have are $( R_i, L_{i-1} )$, since it's easy to see that any legal blue arc covers one of these. If we have exactly one of these for each $i$, the theorem is true because, except for the endpoints of the arcs, every point is covered exactly $n-1$ times, and all the endpoints of the arcs are covered n times.

The only thing remaining is to prove that if we have multiple copies of some $( R_i, L_{i-1} )$, and none of others, the theorem still holds. Looking at examples this seems to be true, but I haven't found a proof.

Added: Sergei Ivanov has a very nice argument that completes the proof; see his answer.

1

Here's the start of a proof.

We will say that two arcs overlap twice if their union covers the entire circle. First, note that you can assume that there are no two red arcs that overlap twice.

Proof: Suppose that there were two blue arcs that overlap twice and two red arcs that overlap twice. Then, we could delete all four of these arcs and find a point covered by $n-2$ of the remaining arcs. Two of the deleted arcs also cover this point, and we're done. By symmetry, we can assume that there are no two red arcs that overlap twice.

Now, note that we can assume that no red arc is contained in another. If there were, we could shrink the larger one so that they overlap, but neither is contained in the other.

These two observations mean that we can order the red arcs clockwise around the circle. Let the red arcs be $(L_i, R_i)$, with $L_i$ being the left (anticlockwise) endpoints and $R_i$ be the right (clockwise) endpoints. Then $L_i$ and $R_i$ are also ordered clockwise around the circle.

Now, we can assume that the only blue arcs we have are $( R_i, L_{i-1} )$, since it's easy to see that any legal blue arc covers one of these. If we have exactly one of these for each $i$, the theorem is true because, except for the endpoints of the arcs, every point is covered exactly $n-1$ times, and all the endpoints of the arcs are covered n times.

The only thing remaining is to prove that if we have multiple copies of some $( R_i, L_{i-1} )$, and none of others, the theorem still holds. Looking at examples this seems to be true, but I haven't found a proof.