In how many ways can $50!$ be expressed as a sum of two or more consecutive positive integers?

The answer is 93000959.

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According the theory for these matters, we need to count the number of odd factors of $50!$ . Each such factor gives rise to a representation of $50!$ as a sum of consecutive positive integers.

Now $50!=2^{47}\times3^{22}\times5^{12}\times7^8\times11^4\times13^3\times17^2\times19^2\times23^2$ $\times29\times31\times37\times41\times43\times47$ . Thus the number of odd factors is $23*13*9*5*4*3^3*2^6=93000960$ . The number of representations we seek is one less, $\boxed{93000959}$ , since we exclude the trivial representation $50!=50!$ .