\(P=\frac{1+2+3+...+105}{1-2+3-4+...+103-104+105}\)
\(=\frac{\frac{105\left(105+1\right)}{2}}{\left(1-2\right)+\left(3-4\right)+...+\left(103-104\right)+105}\)
\(=\frac{\frac{105\left(105+1\right)}{2}}{-1+\left(-1\right)+....+\left(-1\right)+105}\)
\(=\frac{5565}{-52+105}=\frac{5565}{53}=105\)