Since these arithmetic series have the same number of addends, we want their averages to be the same. For any given $n,$ the last addend of $S$ is $n$ and the last addend in $T$ is $102-2n.$ So, the average of the addends in $S$ is $\frac{n+1}2$ and the average of the addends in $T$ is $\frac{100 + (102 - 2n)}2=101-n.$ We easily solve the resulting linear equation: \begin{align*} \frac{n+1}2 &= 101-n \\ \frac{3n+1}2 &= 101 = \frac{202}2 \\ \frac{3n}2 &= \frac{201}2 \\ n &= \boxed{67}. \end{align*}