janos' answer is good.

Let's me add a few simple independant comments on top of his solution :

 - Unlike C or C++, Python's modulo operator (%) always return a number having the same sign as the denominator (divisor). Thus, as D is supposed to be positive, `(Y - X) % D >= 0`. Therefore, your check could be written  `if (Y - X) % D != 0` or in a more Pythonic way : `if (Y - X) % D`.

 - Python has a pretty cool [divmod][1] function. It computes quotient and reminder which is exactly what you want here. Your function becomes :

        def solution(X, Y, D):
            q, r = divmod(Y-X, D)
            if r > 0:
                return q + 1
            return q


 - You could avoid the repeted `return jump` by using the ternary operator : `return jumps + (1 if (Y - X) % D > 0 else 0)`.

By taking all comments into account, your code becomes :

    def solution2(X, Y, D):
        q, r = divmod(Y-X, D)
        return q + (1 if r else 0)

It cannot really get any simpler, can it ?

  [1]: https://docs.python.org/2/library/functions.html#divmod