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We claim this holds exactly for $n$ even.
First solution by generating functions Define
$$R(x) = \sum_{s \text{ red}} x^s, \quad B(x) = \sum_{s \text{ blue}} x^s.$$
(Here “blue” means “not-red”, as always.) Then, the number of tuples in $T_n$ with exactly $k$ red coordinates is exactly equal to
$$\binom{n}{k} \cdot \frac{1}{100} \sum_{\omega} R(\omega)^k B(\omega)^{n-k}$$
where the sum is over all primitive 100th roots of unity. So, we conclude the number of tuples in $T_n$ with an even (resp odd) number of red elements is exactly
$$\begin{align*} X &= \frac{1}{100} \sum_{\omega} \sum_{k \text{ even}} \binom{n}{k} R(\omega)^k B(\omega)^{n-k} \\ Y &= \frac{1}{100} \sum_{\omega} \sum_{k \text{ odd}} \binom{n}{k} R(\omega)^k B(\omega)^{n-k} \\ \implies X - Y &= \frac{1}{100} \sum_{\omega} (B(\omega) - R(\omega))^n \\ &= \frac{1}{100} \left[ (B(1) - R(1))^n + \sum_{\omega \neq 1} (2B(\omega))^n \right] \\ &= \frac{1}{100} \left[ (B(1) - R(1))^n - (2B(1))^n + 2^n \sum_{\omega} B(\omega)^n \right] \\ &= \frac{1}{100} \left[ (B(1) - R(1))^n - (2B(1))^n \right] + 2^n Z \\ &= \frac{1}{100} [(-50)^n - 50^n] + 2^n Z. \end{align*}$$
where
$$Z \stackrel{\text{def}}{=} \frac{1}{100} \sum_{\omega} B(\omega)^n \geq 0$$
counts the number of tuples in $T_n$ which are all blue. Here we have used the fact that $B(\omega) + R(\omega) = 0$ for $\omega \neq 1$.

We wish to show $X - Y \geq 0$ holds for $n$ even, but may fail when $n$ is odd. This follows from two remarks:
* If $n$ is even, then $X - Y = 2^n Z \geq 0$.
* If $n$ is odd, then if we choose the coloring for which $s$ is red if and only if $s \neq 2 \pmod 4$; we thus get $Z = 0$. Then $X - Y = -\frac{2}{100} \cdot 50^n < 0$.

Second solution by strengthened induction and random coloring We again prove that $n$ even work. Let us define
$$T_n(a) = \{(a_1, \dots, a_n) \in S^n \mid a_1 + \dots + a_n \equiv a \pmod{100}\}.$$
Also, call an $n$-tuple good if it has an even number of red elements. We claim that $T_n(a)$ also has at least 50% good tuples, by induction.
This follows by induction on $n \geq 2$. Indeed, the base case $n = 2$ can be checked by hand, since $T_2(a) = \{(x, a-x) \mid x \in S\}$. With the stronger claim, one can check the case $n = 2$ manually and proceed by induction to go from $n-2$ to $n$, noting that
$$T_n(a) = \bigsqcup_{b+c=a} T_{n-2}(b) \oplus T_2(c)$$
where $\oplus$ denotes concatenation of tuples, applied set-wise. The concatenation of an $(n-2)$-tuple and 2-tuple is good if and only if the both or neither are good. Thus for each $b$ and $c$, if the proportion of $T_{n-2}(b)$ which is good is $p \geq \frac{1}{2}$ and the proportion of $T_2(c)$ which is good is $q \geq \frac{1}{2}$, then the proportion of $T_{n-2}(b) \oplus T_2(c)$ which is good is

$pq + (1-p)(1-q) \ge \frac{1}{2}$, as desired. Since each term in the union has at least half the tuples good, all of $T_n(a)$ has at least half the tuples good, as desired.
It remains to fail all odd $n$. We proceed by a suggestion of Yang Liu and Ankan Bhattacharya by showing that if we pick the 75 elements randomly, then any particular tuple in $S^n$ has strictly less than 50% chance of being good. This will imply (by linearity of expectation) that $T_n$ (or indeed any subset of $S^n$) will, for some coloring, have less than half good tuples.
Let $(a_1, \dots, a_n)$ be such an $n$-tuple. If any element appears in the tuple more than once, keep discarding pairs of that element until there are zero or one; this has no effect on the good-ness of the tuple. If we do this, we obtain an $m$-tuple $(b_1, \dots, b_m)$ with no duplicated elements where $m \equiv n \equiv 1 \pmod 2$. Now, the probability that any element is red is $\frac{3}{4}$, so the probability of being good is
$$\sum_{k \text{ even}}^{m} \binom{m}{k} \left(\frac{3}{4}\right)^k \left(-\frac{1}{4}\right)^{m-k} = \frac{1}{2} \left[ \left(\frac{3}{4} + \frac{1}{4}\right)^m - \left(\frac{3}{4} - \frac{1}{4}\right)^m \right] \\ = \frac{1}{2} \left[ 1 - \left(\frac{1}{2}\right)^m \right] < \frac{1}{2}.$$