Let $x_1,x_2,\dots ,x_{100}$ be $100$ numbers such that $x_i+x_{i+1}=100$ for all $i$. If $x_{10}=1$, then the value of $x_1$ is

Given the relationship:
$$x_i+x_{i+1}=100$$

For $i=1$:

$$x_1+x_2=100⟹x_2=100-x_1$$

For $i=2$:

$$x_2+x_3=100⟹(100-x_1)+x_3=100⟹x_3=x_1$$

For $i=3$:

$$x_3+x_4=100⟹x_1+x_4=100⟹x_4=100-x_1$$

By observing this alternating sequence, we find that for any odd index $k$, $x_k=x_1$, and for any even index $k$, $x_k=100-x_1$.
Since $10$ is an even index:

$$x_{10}=100-x_1$$

Given that $x_{10}=1$:

$$1=100-x_1⟹x_1=99$$