Show it is impossible using Chinese Remainder Theorem $n = pq$, where $p$, $q$ are primes.

Let $m$ be a universal exponent, i.e. $\,a^m \equiv 1 \pmod n\ $ for all $a$ coprime to $n$

Then there are four cases:
\begin{cases} a^{m/2} \equiv\ \ \ 1\ \pmod p\\ 
a^{m/2} \equiv\ \ \ 1\ \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv\ \ \ 1 \pmod p\\ 
a^{m/2} \equiv -1 \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\ 
a^{m/2} \equiv\ \ \ 1\ \pmod{\! q} \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\ 
a^{m/2} \equiv -1 \pmod q \end{cases}

My question is that, for all $a$ such that $\gcd(a,n)=1$, why can't half of '$a$'s satisfy the first case and half satisfy the last case.( i.2. show it impossible to only have case 1 and 4).

I think I can use Chinese Remainder Theorem to solve it.

It gives exactly half of '$a$'s satisfy $a^{m/2} \equiv -1 \pmod n$ and the rest satisfy $a^{m/2} \equiv 1 \pmod n$. But I still didn't get why it is impossible.
Assume $p$ and $q$ are odd, distinct primes (if one of the primes is $2$, then $1 \equiv -1 \pmod 2$, so the question becomes somewhat meaningless in that case). Your question can then be proven using contradiction.

Assume that only both case $1$ and case $4$ apply, i.e.,

\begin{cases} a^{m/2} \equiv\ \ \ 1\ \pmod p\\ 
a^{m/2} \equiv\ \ \ 1\ \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\ 
a^{m/2} \equiv -1 \pmod q \end{cases}

Consider all $a_b = bp + 1$ for $0 \le b \le q - 1$. First, note $a_b^{m/2} \equiv 1 \pmod p$ for all $a_b$. Thus, by the assumption, $a_b^{m/2} \equiv 1 \pmod q$ for all these $a_b$ which are relatively prime to $q$. However, note that modulo $q$, the set of $a_b$ covers all values from $0$ to $q-1$, inclusive, because, if we have $b_1 p + 1 \equiv b_2 p + 1 \pmod q$, then $(b_1 - b_2)p \equiv 0 \pmod q \implies b_1 - b_2 \equiv 0 \pmod q \implies b_1 = b_2$, since $p$ and $q$ are distinct primes, for any $0 \le b_1, b_2 \le q - 1$. Thus, this means that $c^{m/2} \equiv 1 \pmod q$ for all $1 \le c \le q - 1$. As such, there are no values of any $a$ where $a^{m/2} \equiv -1 \pmod q$, unless $1 \equiv -1 \pmod q$, but this is only true if $q = 2$, which is not allowed as $q$ is an odd prime. This shows only case $1$ would apply and, thus, you can't have just both cases $1$ and $4$ apply.

As for using the Chinese Remainder Theorem to prove this, note this theorem is really only an existence theorem, so I'm not sure how you could use it. If you perhaps meant using something like the Extended Euclidean Algorithm, note that just by inspection you can see the question comes down to trying to prove $a^{m/2} = kn + 1$ for some $k \in \mathbb{Z}$ (i.e., $a^{m/2} \equiv 1 \pmod n$), or $a^{m/2} = kn - 1$ for some $k \in \mathbb{Z}$, (i.e., $a^{m/2} \equiv -1 \pmod n$), with at least one case of each, for all $a$ relatively prime to $p$ and $q$. If this is what you mean, you could use basically the same proof technique I used, although I believe the notation &amp; handling would not be quite as convenient.
Assume $p$ and $q$ are odd, distinct primes (if one of the primes is $2$, then $1 \equiv -1 \pmod 2$, so the question becomes somewhat meaningless in that case). Your question can then be proven using contradiction.

Assume that only both case $1$ and case $4$ apply, i.e.,

\begin{cases} a^{m/2} \equiv\ \ \ 1\ \pmod p\\ 
a^{m/2} \equiv\ \ \ 1\ \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\ 
a^{m/2} \equiv -1 \pmod q \end{cases}

Consider all $a_b = bp + 1$ for $0 \le b \le q - 1$. First, note $a_b^{m/2} \equiv 1 \pmod p$ for all $a_b$. Thus, by the assumption, $a_b^{m/2} \equiv 1 \pmod q$ for all these $a_b$ which are relatively prime to $q$. However, note that modulo $q$, the set of $a_b$ covers all values from $0$ to $q-1$, inclusive, because, if we have $b_1 p + 1 \equiv b_2 p + 1 \pmod q$, then $(b_1 - b_2)p \equiv 0 \pmod q \implies b_1 - b_2 \equiv 0 \pmod q \implies b_1 = b_2$, since $p$ and $q$ are distinct primes, for any $0 \le b_1, b_2 \le q - 1$. Thus, this means that $c^{m/2} \equiv 1 \pmod q$ for all $1 \le c \le q - 1$. As such, there are no values of any $a$ where $a^{m/2} \equiv -1 \pmod q$, unless $1 \equiv -1 \pmod q$, but this is only true if $q = 2$, which is not allowed as $q$ is an odd prime. This shows only case $1$ would apply and, thus, you can't have just both cases $1$ and $4$ apply.

As for using the Chinese Remainder Theorem to prove this, note this theorem is really only an existence theorem, so I'm not sure how you could use it. If you perhaps meant using something like the Extended Euclidean Algorithm, note that just by inspection you can see the question comes down to trying to prove $a^{m/2} = kn + 1$ for some $k \in \mathbb{Z}$ (i.e., $a^{m/2} \equiv 1 \pmod n$), or $a^{m/2} = kn - 1$ for some $k \in \mathbb{Z}$, (i.e., $a^{m/2} \equiv -1 \pmod n$), with at least one case of each, for all $a$ relatively prime to $p$ and $q$. If this is what you mean, you could use basically the same proof technique I used, although I believe the notation &amp; handling would not be quite as convenient.