user411: As you know, in the initial position Chess960 the king lies between the two rooks. Here are the rules for castling.
For a starting position P, let n(P) be the minimal number of moves that White needs to legally castle kingside. How many positions are there such that n(P) = 1, n(P) = 2, and so on?
There are cases where we can castle in the first move, namely when the king on f1 and the rook on g1. There are also other extreme cases with the king on b1 and the rook on c1, so then we have to make space.
Maybe some engines can solve this question.

user14079: Edit: This answer only deals with kingside castling; the figures have to be revised if you also consider queenside castling.
The three first steps:
There are 90 positions where n(P)=1
For White to be able to castle short at once, he needs to have Kf1, Rg1 in the initial position. Then there are 5 possible spots from the second rook, 3 for one bishop and 2 for the other one, and finally 3 remaining spots for the queen. The knights go to the last two spots.
There are 100 positions where n(P)=2
36 of them with Kf1, Ng1, Rh1: White can play 1.Nf3 or 1.Nh3 and 2.0-0
36 of them with Kg1, Nf1, Rh1: White can play 1.Ne3 or 1.Ng3 and 2.0-0
28 of them with Rg1, Nf1, Ke1: White can play 1.Ng3 or 1.Ne3 and 2.0-0
There are 236 positions where n(P)=3
please double-check this part
Here are the different configurations and the number of relevant positions:
xxxxxQKR 18
xxxxxBKR 36
xxxxxRKR 18
xxxxxKQR 18
xxxxxKBR 24
xxxxKNNR 8
xxxKNNRx 6
xxxKNRNx 6
xxxKRNNx 6
xxxxKBRx 24
xxxxKQRx 18
xxxxKRBx 36
xxxxKRQx 18
Total:236
Edit : removed 2 cases after checking the rules: the path to g1 must be open for the king (only the right-hand rook can be in the way)
Once we reach n(P)=4, we cannot ignore Black's play anymore: for instance, QNRNKBBR looks like it should have n=5, but actually it has n=4 because of the sequence 1.f4 b5 2.Bxa7 Qxg2 3.Bxg2 Nc6 4.0-0. Counting by hand becomes too fastidious and hazardous: a software program is needed.

On the other hand, after clarification by the OP, I believe 7 to be the maximum value of the function P->n(P).
If Black wasn't allowed to "help" we could build only 2 positions with n(P)=9; Here is one, the other is similar with RKBBNNQR on the first rank.
[FEN "rkqnnbbr/pppppppp/8/8/8/8/PPPPPPPP/RKQNNBBR w - - 0 1"]

In a series problem where White plays alone he would need 9 moves in order to castle. But since Black moves are taken into account, White actually can castle on move 7 from that diagram if the game goes something like 1.b3 d5 2.Qa3 Qe6 3.Nb2 Qxe2 4.Ned3 Qxf2 5.Bxf2 Nc6 6.Be2 0-0-0 7.0-0.

411: <p>As you know, in the initial position Chess960 the king lies between the two rooks. <a href="https://chess960.net/how-to-play/" rel="nofollow noreferrer">Here</a> are the rules for castling.</p>
<p>For a starting position <em>P</em>, let <em>n</em>(<em>P</em>) be the minimal number of moves that White needs to legally castle kingside. How many positions are there such that <em>n</em>(<em>P</em>) = 1, <em>n</em>(<em>P</em>) = 2, and so on?</p>
<p>There are cases where we can castle in the first move, namely when the king on f1 and the rook on g1. There are also other extreme cases with the king on b1 and the rook on c1, so then we have to make space.</p>
<p>Maybe some engines can solve this question.</p>
 14079: <p>Edit: This answer only deals with kingside castling; the figures have to be revised if you also consider queenside castling.</p>
<p>The three first steps:</p>
<p><strong>There are 90 positions where n(P)=1</strong></p>
<p>For White to be able to castle short at once, he needs to have Kf1, Rg1 in the initial position. Then there are 5 possible spots from the second rook, 3 for one bishop and 2 for the other one, and finally 3 remaining spots for the queen. The knights go to the last two spots.</p>
<p><strong>There are 100 positions where n(P)=2</strong></p>
<p>36 of them with Kf1, Ng1, Rh1: White can play 1.Nf3 or 1.Nh3 and 2.0-0</p>
<p>36 of them with Kg1, Nf1, Rh1: White can play 1.Ne3 or 1.Ng3 and 2.0-0</p>
<p>28 of them with Rg1, Nf1, Ke1: White can play 1.Ng3 or 1.Ne3 and 2.0-0</p>
<p><strong>There are 236 positions where n(P)=3</strong>
<em>please double-check this part</em></p>
<p>Here are the different configurations and the number of relevant positions:</p>
<p>xxxxxQKR 18<br>
xxxxxBKR 36<br>
xxxxxRKR 18<br>
xxxxxKQR 18<br>
xxxxxKBR 24<br>
xxxxKNNR 8<br>
xxxKNNRx 6<br>
xxxKNRNx 6<br>
xxxKRNNx 6<br>
xxxxKBRx 24<br>
xxxxKQRx 18<br>
xxxxKRBx 36<br>
xxxxKRQx 18<br></p>
<p>Total:236</p>
<p><em>Edit : removed 2 cases after checking the <a href="https://chess960.net/how-to-play/" rel="nofollow noreferrer">rules</a>: the path to g1 must be open for the king (only the right-hand rook can be in the way)</em></p>
<p>Once we reach <strong>n(P)=4</strong>, we cannot ignore Black's play anymore: for instance, QNRNKBBR <em>looks like</em> it should have n=5, but actually it has n=4 because of the sequence 1.f4 b5 2.Bxa7 Qxg2 3.Bxg2 Nc6 4.0-0. Counting by hand becomes too fastidious and hazardous: a software program is needed.</p>
<hr />
<p>On the other hand, after clarification by the OP, I believe <strong>7 to be the maximum</strong> value of the function P-&gt;n(P).</p>
<p>If Black wasn't allowed to &quot;help&quot; we could build only 2 positions with n(P)=9; Here is one, the other is similar with RKBBNNQR on the first rank.</p>
<pre><code>[FEN &quot;rkqnnbbr/pppppppp/8/8/8/8/PPPPPPPP/RKQNNBBR w - - 0 1&quot;]
</code></pre>
<p>In a series problem where White plays alone he would need 9 moves in order to castle. But since Black moves are taken into account, White actually can castle on move 7 from that diagram if the game goes something like 1.b3 d5 2.Qa3 Qe6 3.Nb2 Qxe2 4.Ned3 Qxf2 5.Bxf2 Nc6 6.Be2 0-0-0 7.0-0.</p>
