We need to find a real number $t$ such that for any set of 120 points $P_1, \ldots, P_{120}$ on the boundary of a unit square, there exists a point $Q$ on this boundary with $|P_1Q| + |P_2Q| + \cdots + |P_{120}Q| = t$.

Define $\mathcal{U}$ to be a set of points $P_1, \ldots, P_{120}$ on the boundary of a unit square. Define $g_{\mathcal{U}}(Q) = \sum_{i=1}^{120} |QP_i|$.

Lemma 1: The set $\{g_{\mathcal{U}}(Q) : Q \in \mathcal{U}\}$ is a closed interval $I_{\mathcal{U}}$.

*Proof:* Clearly, $g_{\mathcal{U}}(Q)$ is bounded above and below over $Q \in \mathcal{U}$, and it is continuous in both $x$ and $y$ coordinates if we place it in the Cartesian plane. Combining these two implies the set of values is an interval. $\blacksquare$

Lemma 2: Given a finite set of closed intervals, they all intersect if and only if every two intersect.

We want to show that the intervals $I_{\mathcal{U}}$ all intersect over all sets of 120 points $\mathcal{U}$. By Lemma 2, it suffices to check that every two intersect. Suppose for the sake of contradiction that there exists some $\mathcal{U} = \{P_1, \ldots, P_{120}\}$ and $\mathcal{U}' = \{P_1', \ldots, P_{120}'\}$ such that $I_{\mathcal{U}}$ is entirely before $I_{\mathcal{U}'}$. The key is that now
$$g_{\mathcal{U}}(Q) < g_{\mathcal{U}'}(Q') \quad \text{for all } Q \in \mathcal{U} \text{ and } Q' \in \mathcal{U}' \quad (\spadesuit).$$
Let $C_1, C_2, C_3, C_4$ be the corners of the unit square $\mathcal{U}$ and $M_1', M_2', M_3', M_4'$ the midpoints of the four sides of the unit square $\mathcal{U}'$. Summing four bounds appearing from $(\spadesuit)$:
$$g_{\mathcal{U}}(C_1) + \cdots + g_{\mathcal{U}}(C_4) < g_{\mathcal{U}'}(M_1) + \cdots + g_{\mathcal{U}'}(M_4) \quad (\clubsuit).$$
The key is that we can compute and bound each of the above since they become sums of functions of a single point $P_i$ relative to the fixed unit square, instead of about the entire set of $P_i$'s. In particular,
$$\begin{align*} g_{\mathcal{U}}(C_1) + \cdots + g_{\mathcal{U}}(C_4) &= \sum_{j=1}^4 \sum_{i=1}^{120} |C_jP_i| \\ &= \sum_{i=1}^{120} |C_1P_i| + |C_2P_i| + |C_3P_i| + |C_4P_i| \\ &\ge \sum_{i=1}^{120} (1 + \sqrt{5}) \\ &= 120(1 + \sqrt{5}). \end{align*}$$
The second step above followed by switching the order of summation. The third step since we can confirm with coordinates that the minimum $|C_1P| + |C_2P| + |C_3P| + |C_4P|$ over $P$ on the boundary occurs is $1 + \sqrt{5}$, and occurs when $P$ is the midpoint of a side. Now similarly,
$$\begin{align*} g_{\mathcal{U}}(M_1') + \cdots + g_{\mathcal{U}}(M_4') &= \sum_{j=1}^4 \sum_{i=1}^{120} |M_j'P_i'| \\ &= \sum_{i=1}^{120} |M_1'P_i'| + |M_2'P_i'| + |M_3'P_i'| + |M_4'P_i'| \\ &\le \sum_{i=1}^{120} (1 + \sqrt{5}) \\ &= 120(1 + \sqrt{5}). \end{align*}$$
The third step since we can confirm with coordinates that the maximum $|M_1P| + |M_2P| + |M_3P| + |M_4P|$ over $P$ on the boundary is $1 + \sqrt{5}$, and occurs when $P$ is a corner.

However, combining these two bounds contradicts $(\clubsuit)$! Therefore, such a $t$ exists. In particular, we can show $t = 30(1 + \sqrt{5})$ by proving that $t < 30(1 + \sqrt{5})$ fails from the corners bound and $t > 30(1 + \sqrt{5})$ fails from the midpoints bound; now, since we have shown at least one valid $t$ exists, it must be the claimed value.

The answer is: $\boxed{30(1 + \sqrt{5})}$.