Consider all the real sequences $x_0,x_1,\cdots,x_{100}$ satisfying the following two requirements:
(1)$x_0=0$;
(2)For any integer $i,1\leq i\leq 100$,we have $1\leq x_i-x_{i-1}\leq 2$.
Find the greatest positive integer $k\leq 100$,so that for any sequence $x_0,x_1,\cdots,x_{100}$ like this,we have
\[x_k+x_{k+1}+\cdots+x_{100}\geq x_0+x_1+\cdots+x_{k-1}.\]
1. We start by considering the given conditions for the sequence \( x_0, x_1, \ldots, x_{100} \):
   - \( x_0 = 0 \)
   - For any integer \( i \) where \( 1 \leq i \leq 100 \), we have \( 1 \leq x_i - x_{i-1} \leq 2 \).

2. We need to find the greatest positive integer \( k \leq 100 \) such that:
   \[
   x_k + x_{k+1} + \cdots + x_{100} \geq x_0 + x_1 + \cdots + x_{k-1}.
   \]

3. Let's generalize the problem by replacing 100 with \( n \). We need to find \( k \) such that:
   \[
   x_k + x_{k+1} + \cdots + x_n \geq x_0 + x_1 + \cdots + x_{k-1}.
   \]

4. By the given conditions, we know that \( x_{j+d} \geq x_j + d \) for all \( j, d \geq 1 \). This implies that the sequence grows at least linearly.

5. We can write:
   \[
   x_k + x_{k+1} + \cdots + x_n \geq x_{k-1} + x_{k-2} + \cdots + x_{2k-1-n} + (n+1-k)^2.
   \]

6. We need to ensure that:
   \[
   x_1 + x_2 + \cdots + x_{2k-2-n} \leq (n+1-k)^2.
   \]

7. The left-hand side (LHS) can be at most:
   \[
   2 + 4 + \cdots + 2(2k-2-n) = (2k-2-n)(2k-1-n).
   \]

8. We need to check if:
   \[
   (2k-2-n)(2k-1-n) \leq (n+1-k)^2.
   \]

9. Simplifying the inequality:
   \[
   (2k-2-n)(2k-1-n) \leq (n+1-k)^2.
   \]

10. Expanding both sides:
    \[
    (2k-2-n)(2k-1-n) = 4k^2 - 4k - 2kn + n^2 - 2k + n.
    \]
    \[
    (n+1-k)^2 = n^2 + 2n + 1 - 2nk - 2k + k^2.
    \]

11. Comparing the coefficients, we need:
    \[
    4k^2 - 4k - 2kn + n^2 - 2k + n \leq n^2 + 2n + 1 - 2nk - 2k + k^2.
    \]

12. Simplifying further, we get:
    \[
    3k^2 - 2kn - 2k \leq 2n + 1.
    \]

13. Solving for \( k \):
    \[
    3k^2 - 2kn - 2k - 2n - 1 \leq 0.
    \]

14. This inequality is satisfied if:
    \[
    k \leq \frac{2n+2}{3}.
    \]

15. Substituting \( n = 100 \):
    \[
    k \leq \frac{2 \cdot 100 + 2}{3} = \frac{202}{3} \approx 67.33.
    \]

16. Therefore, the greatest integer \( k \) is \( 67 \).

The final answer is \( \boxed{67} \).