Proof:
Let the integers be $\mathrm {\color {blue} {x_1, x_2, x_3,\cdots \cdots , x_{105}}}$
Assume the following sums:
$\mathrm {S_1 = x_1}$
$\mathrm {S_2 = x_1 + x_2}$
$\vdots \;\;\;= \;\;\; \vdots $
$\vdots \;\;\;= \;\;\; \vdots $
$\mathrm {S_k = x_1 + x_2+\cdots +x_k}$
Thus, in total we have $\mathrm {S_1, S_2, \cdots S_{105}}$ terms. Also, we know that in Modulo $99$, there can be only $99$ possible remainders.
This proves that there has to be some $\mathbf {i<j}$ where $\mathrm {S_i \equiv S_j (\text{mod} \;99)}$
$\therefore $ $99$ divides $\mathrm {S_j – S_i = x_{i+1} + x_{i+2}+...+x_j}$