On a game with cuisenaire rods

Here is a variation of a Nim game : consider a full set of cuisenaire rods - 10 rods of all integer lengths between 1 and 10. Set a number N between 1 and 54. Player 1 choose one of the 10 rods and place it between the two players. Player 2 choose one of the remaining 9 rods and put it following the first one. Player 1 choose one of the remaining 8 rods and place it following the second one, and so on. The loser of the game is the player that place a rod such that the sum of the lengths is greater than N.

Quite strangely, determining whether player 1 wins or loses may be very easy or quite difficult with respect to N. Do you know anything about this game? I can't say that I invented it as I am sure that someone else has already studied it but I don't know where to search. Thanks by advance for your comments :)

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Where do the $44$ come from? – Christian Blatter Jul 29 '13 at 15:17
I think it's supposed to be $54$. $1+2+3+\ldots+10=55$, so for $N>54$ there's no win condition. – Daniel Franke Jul 29 '13 at 15:44
Sorry, the upper bound is actually 54 – Julien Jul 29 '13 at 15:48
This looks hard in general, because a position depends not only on the sum of the lengths already played, but on the exact subset of rods that remain unplayed. But with only ten rods, there are just 1024 possible positions, which could easily be handled by a computer search. – TonyK Jul 29 '13 at 15:54
Of course this is a finite game with not so many possibilities for a computer, but it seems to be not so easy for a human player to deal with those... – Julien Jul 29 '13 at 16:56

Clearly the first player wins any game with $N \lt 11$. The second player wins $11, 12$ ($12$ by playing $10$ if the first player starts with $1$). First wins $13,14$ by playing $1$.
First wins $54$. Second wins $53$ by playing $1$ and $52,51$ by playing $1,2,3$