Given the functions $f(x)=\sin(2x+\frac{\pi}{4})$ and $g(x)=\sin(2x+\frac{3\pi}{4})$, we are asked to find the maximum value of the positive real number $t$ for which $f(x_1) - f(x_2) < g(x_1) - g(x_2)$ always holds for $0 \leqslant x_1 < x_2 \leqslant t$.

First, we observe that the inequality $f(x_1) - f(x_2) < g(x_1) - g(x_2)$ can be rewritten as:
$$f(x_1) - g(x_1) < f(x_2) - g(x_2).$$
This suggests defining a new function $h(x) = f(x) - g(x)$, which implies that $h(x)$ is monotonically increasing on the interval $[0, t]$.

Next, we calculate $h(x)$:
\begin{align*}
h(x) &= \sin(2x+\frac{\pi}{4}) - \sin(2x+\frac{3\pi}{4}) \\
&= \sin(2x+\frac{\pi}{4}) - \sin\left(\frac{\pi}{2} + (2x+\frac{\pi}{4})\right) \\
&= \sin(2x+\frac{\pi}{4}) - \cos(2x+\frac{\pi}{4}) \\
&= \sqrt{2}\left(\frac{\sqrt{2}}{2}\sin(2x+\frac{\pi}{4}) - \frac{\sqrt{2}}{2}\cos(2x+\frac{\pi}{4})\right) \\
&= \sqrt{2}\sin(2x).
\end{align*}

The function $\sin(2x)$ is monotonically increasing in the intervals where $-\frac{\pi}{2} + 2k\pi \leq 2x \leq \frac{\pi}{2} + 2k\pi$, for $k \in \mathbb{Z}$. This translates to intervals of $x$ as $-\frac{\pi}{4} + k\pi \leq x \leq \frac{\pi}{4} + k\pi$, for $k \in \mathbb{Z}$.

For $k=0$, the interval where $h(x)$ is monotonically increasing is $[-\frac{\pi}{4}, \frac{\pi}{4}]$. Beyond this interval, specifically in $(\frac{\pi}{4}, \frac{3\pi}{4})$, $h(x)$ starts to decrease.

Therefore, to ensure $h(x)$ is monotonically increasing on the interval $[0, t]$, the maximum value of $t$ must be $\frac{\pi}{4}$.

Hence, the maximum value of the positive real number $t$ for which the given condition holds is $\boxed{\frac{\pi}{4}}$.