I will prove $t= 30(1+\sqrt{5})$ is the answer ( counterex. are possible for other values). We first prove for every $120$ points we can choose a $Q$ such that the sum is max. $t:$ We call the unit square ABCD and the midpoints M,N,O,P. So we get AMBNCODP. Take $Q=M,N,O,P$ resp., then the four given sums are together equal to $120$ sums of form $|P_1M|+|P_1O|+|P_1N|+|P_1P|.$ But each such sum is between $1+\sqrt{2}$ and $1+\sqrt{5}$ so one of the 4 possible ways for Q satisfies. Similar for $Q=A,B,C,D$ we get for every configuration there is a point $Q$ such that the sum is min. $t.$

SCP wrote: We first prove for every $120$ points we can choose a $Q$ such that the sum is max. $t:$ We call the unit square ABCD and the midpoints M,N,O,P. So we get AMBNCODP. Take $Q=M,N,O,P$ resp., then the four given sums are together equal to $120$ sums of form $|P_1M|+|P_1O|+|P_1N|+|P_1P|.$ But each such sum is between $1+\sqrt{2}$ and $1+\sqrt{5}$ so one of the 4 possible ways for Q satisfies. Similar for $Q=A,B,C,D$ we get for every configuration there is a point $Q$ such that the sum is min. $t.$ What do you mean about the maximum and minimum of t?? I can't understand. Could you explain with more details? Thanks.

Here is an approach which only shows some $t$ exists (it is then possible to show that $t = 30(1 + \sqrt{5})$ is the only valid one without much difficulty). Let $S$ be the boundary. For any sequence $A = A_1, \dots, A_{120}$ on the boundary, define \[\alpha(A) = \{A_1P + \dots + A_{120}P : P \in S\}.\]Since $S$ is connected and compact it follows every $\alpha(A)$ is a closed interval. Thus to show that all $\alpha(A)$ intersect, it is enough to show that any two, say $\alpha(A)$ and $\alpha(B)$, do. Suppose not, and WLOG assume \[A_1P + \dots + A_{120}P < B_1Q + \dots + B_{120}Q\]for all $P, Q \in S$. Selecting the 120 pairs $(P, Q) = (B_k, A_k)$ and summing we obtain \[\sum_{i = 1}^{120} \sum_{j = 1}^{120} A_i B_j < \sum_{i = 1}^{120} \sum_{j = 1}^{120} B_i A_j,\]contradiction. Thus any two $\alpha(A)$ intersect, and so all of them do.

Let's clear up the SCP's solution. Very nice idea. Notations: $ABCD$ be the unit square, $M,N,P,O$ be the midpoints of its sides, $S(X):=\sum_{i=1}^n |P_iX|$. For brevity put $n$ instead $120$. Let $P_1,P_2,\dots,P_n$ be some fixed $n$ points on the square's boundary. It can be easily checked: $$|P_iA|+|P_iB|+|P_iC|+|P_iD|\ge 1+\sqrt{5}$$Summing on $i=1,\dots,n$, we get: $$S(A)+S(B)+S(C)+S(D)\ge (1+\sqrt{5})n $$It means there exists $X$ on the boundary of square for which $S(X)\ge (1+\sqrt{5})n/4$, for example one of the points $A,B,C,D$. Further, it can be checked: $$|P_iM|+|P_iN|+|P_iP|+|P_iO|\le 1+\sqrt{5}$$Summing, we get: $$S(M)+S(N)+S(P)+S(O)\le (1+\sqrt{5})n $$Hence, there exists $Y$ on the boundary, for which $S(Y)\le (1+\sqrt{5})n/4$. Finally, since $S(X)$ is continuous, there exists a point $Q$ on the boundary of $ABCD$, for which: $$S(Q)= (1+\sqrt{5})n/4.$$

@CantonMathGuy: Nice approach! CantonMathGuy wrote: ...it is then possible to show that $t = 30(1 + \sqrt{5})$ is the only valid one without much difficulty... but I think, it's not only one $t$ satisfying conditions. There is an interval of such $t$'s

dgrozev wrote: There is an interval of such $t$'s Are you sure? I think that taking $30 \times \{A, B, C, D\}$ shows $t < 30(1 + \sqrt{5})$ fail, and $30 \times \{M, N, O, P\}$ show $t > 30(1 + \sqrt{5})$ fail.

You're right, my bad.

Extremely nice problem. Local-to-global kind of thinking. Define $\mathcal{U}$ to be a set of points $P_1,\ldots,P_{120}$ on the boundary of a unit square. Define $g_{\mathcal{U}}(Q)=\textstyle \sum_{i=1}^{120}|QP_i|$. Lemma 1: The set $\{g_{\mathcal{U}}(Q):Q\in U\}$ is an (closed) interval $I_{\mathcal{U}}$. Proof: Clearly $g_{\mathcal{U}}(Q)$ is bounded above and below over $Q\in \mathcal{U}$, and also 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: [Well-known] 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 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+\sqrt5 \\ &= 120(1+\sqrt5). \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 $1+\sqrt5$, 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+\sqrt5 \\ &= 120(1+\sqrt5). \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 $1+\sqrt5$, and occurs when $P$ is a corner. However, combining these two bounds contradicts $(\clubsuit)$! Therefore, such a $t$ exists. In particular, we can show $t=30(1+\sqrt5)$ by proving that $t<30(1+\sqrt5)$ fails from the corners bound and $t>30(1+\sqrt5)$ fails from the midpoints bound; now, since we have shown at least one valid $t$ exists, it must be the claimed value.

We claim that for all $n=4k$, the answer is $(1+\sqrt{5})\cdot k$. Let the unit cube be $A=(0,0),B=(1,0),C=(1,1),D=(0,1)$. Then, let the midpoints be $M\in AB,N\in BC,R\in CD,S\in DA$. We use two claims. Claim 1: . For any point $Q$ \[|QA|+|QB|+|QC|+|QD| \geq 1+\sqrt{5}\]Proof: . Assume $Q\in AB$. Then, if we let $C'$ be the reflection of $C$ over $AB$. Then, the expression equals \[|QA|+|QB|+|QC|+|QD| = 1 + |DQ|+|QC'| \geq 1+|DC'| = 1+\sqrt{5}\]$\square$. Claim 1: . For any point $Q$ \[|QM|+|QN|+|QR|+|QS| \leq 1+\sqrt{5}\] Proof: . Assume $Q\in AB$. Firstly, $|QM|+|QS|\leq |QM|+|QA|+|AS| = 1$. Also, both $|QM|,|QN| \leq \frac{\sqrt{5}}{2}$. Thus, combined, the sum is $\leq 1+\sqrt{5}$ $\square$. We now prove that $t = (1+\sqrt{5})k$ works. Firstly, we can get a $Q$ with sum of distances at least $(1+\sqrt{5})k$ by Claim 1 and pigeonhole on $Q\in \{M,N,R,S\}$. Secondly, we can get a $Q$ with sum of distances at most $(1+\sqrt{5})k$ by pigeonhole on Claim 2 choosing $Q\in \{A,B,C,D\}$. Thus, by continuity of the sum, we can achieve $(1+\sqrt{5})k$, and we're done. $\blacksquare$. Remark: t is also clearly unique!

this problem is sicc Given a fixed set $\mathcal{P}$ of points $P_i$ on the boundary, define $I_P$ to be the set of values that is attained by $f(Q) = \sum |QP_i|$ as $Q$ varies on the boundary. It is clear that $I_P$ should always be a singular interval since this function $f$ is continuous. It suffices to show $I_A, I_B$ intersect for any sets $A = \{A_i\}, B = \{B_i\}$. FTSoC that is not the case. WLOG assume $\sum |QA_i| < \sum |RB_i|$ for any $(Q, R)$ on the boundary. Consider the $120$ pairs $(Q, R) = (B_k, A_k)$. Summing\[\sum_i |B_kA_i| < \sum_i |A_kB_i|\]across all $k$ yields\[\sum_{i, k} |B_kA_i| < \sum_{i, k} |A_kB_i|\]which is a contradiction since these sums are the same; they sum the lengths of the same $120^2$ segments. Hence, any two intervals $I_A, I_B$ intersect, so there is a real value in all such intervals, and is thus achievable by $f(Q)$ for any set of points $P$, as desired.

Sketch: Using the fact that the sum of the lengths of any point on the perimeter to the four corners is at least $1+\sqrt{5}$, a switching summation/pigeonhole argument gives that there exists a corner $C$ with $\sum_{i=1}^{120} |P_iC| \ge 30+30\sqrt{5}$. On the other hand, as the sum of the lengths of any point on the perimeter to the four midpoints of the sides is at most $1+\sqrt{5}$, a similar argument concludes that there exists a midpoint $M$ with $\sum_{i=1}^{120} |P_iM| \le 30+30\sqrt{5}$. IVT yields the desired result.

We claim that $t=30+30\sqrt{5}$ works. Note that since it is continous, by IVT it sufficies to show that it is always possible to obtain a distance of at least $30+30\sqrt{5}$ and that it is always possible to obtain a distance of at most $30+30\sqrt{5}.$ First, we will show the "at least" part. We claim that a value of at least $30+30\sqrt{5}$ is obtained by at least one of the vertices of the square. Consider one of the points $P_i$ and analyze the sum of the distances from $P_i$ to the vertices. Two of the distances will have sum 1, and the other two will be $\sqrt{1+r^2}$ and $\sqrt{1+(1-r)^2}$ for some $0\leq r\leq 1.$ Therefore, the sum of the distances from $P_i$ to the vertices is $$1+\sqrt{1+r^2}+\sqrt{1+(1-r)^2}.$$Since $f(x)=\sqrt{1+x^2}$ is convex, by Jensen's, we have $$1+f(r)+f(1-r)\geq 1+2f(1/2)=1+\sqrt{5}.$$Summing this over all $P_i$, the sum of the values of all four vertices is at least $120+120\sqrt{5}$. Thus, at least one vertex must have a value of at least $30+30\sqrt{5}$, which shows this part. Now, let's show the "at most" part. We claim that a value of at most $30+30\sqrt{5}$ is achieved by at least one of the midpoints of the sides. Let $A,B,C,D$ be the vertices, and let $M_1,M_2,M_3,M_4$ be the midpoints of $AB,BC,CD,DA$. Consider a point $P$ on the square. We will show that $PM_1+PM_2+PM_3+PM_4\leq 1+\sqrt{5}.$ By symmetry, WLOG that $P$ is on $AM_1$. Consider what happens as $P$ moves from $M_1$ to $A$. It is moving away from $M_1,M_2,M_3$, but it is moving towards $M_4$. However, the distance will still increase since it is moving away directly from $M_1$ but indirectly towards $M_4$. Therefore, the maximum distance sum is achieved at $A$, giving a distance of $1+\sqrt{5}$. Summing this over all 120 points, the sum of the values of the $M_1,M_2,M_3,M_4$ is at most $120+120\sqrt{5}$, hence one of them has at most $30+30\sqrt{5}.$ Since it is always possible to obtain a distance of at least $30+30\sqrt{5}$ and it is always possible to obtain a distance of at most $30+30\sqrt{5},$ by IVT we can always obtain $30+30\sqrt{5}$, QED.

Here's a sketch of a solution: Let the $4$ vertices be $ABCD$. Then for any given point $P_i$, we can bound the sum $AP_i + BP_i + CP_i + DP_i$. Notice that this sum is at least $1 + \sqrt{5}$ so $\sum_i AP_i + BP_i + CP_i + DP_i \geq 120 + 120\sqrt5$ so there is some vertex of $ABCD$, $X$ so that $\sum_i XP_i \geq 30 + 30\sqrt5$. Then similarly given midpoints $M_1M_2M_3M_4$ we have that there is some $M_j$ so that $\sum_i M_jP_i \leq 30 + 30\sqrt{5}$. So from IVT we can now find a point on the boundary that satisfies $t = 30 + 30\sqrt5$.

Let the square be $ABCD.$ Then for a $P_i,$ we claim that the sum of the distances from $P_i$ to $A,B,C,D$ is $\geq 1+\sqrt{5},$ achieved by having $P_i$ be a midpoint of a side. Consider the ellipse with foci $(-\frac 12, 1), (\frac 12, 1).$ Then the smallest sum of distances where the ellipse goes through the $x$ axis must go through $(0,0).$ Therefore there must be some vertice of the square such that the sum of the lengths is $\geq 30+30\sqrt{5}.$ Similarly, we claim that there must be a midpoint of a side such that the sum of the lengths to that midpoint is $\leq 30+30\sqrt{5}.$ Note that summing over all midpoints and dividing by four is the same as summing over one midpoint however each $P$ is summed four times, each on a different side of the square where the $P$ is rotated about the center of the square. We can see that the maximum is when a $P$ is on the vertice of the square by observing that when we move $P$ towards a midpoint the distance must get smaller because $\sqrt{x^{2}+\frac{1}{4}}+\frac{1}{2}-x$ is decreasing. This means that there must be a midpoint such that the sum of distance is $\leq 30+30\sqrt{5},$ so our answer is $\boxed{30+30\sqrt{5}}.\blacksquare$

In fact, the problem still holds true for any regular $n$-gon $\mathcal P$ and any number of points chosen on the perimeter. Define the distance sum of any point $Q$ as $P_1Q + P_2Q + \dots + P_k Q$. Let $S$ be the distance sum from a midpoint of a side of $\mathcal P$ to the vertices of $\mathcal P$. Due to IVT, it suffices to find a point $Q$ on the perimeter of $\mathcal P$ with distance sum at least $\tfrac{Sk}{n}$, and a point $Q$ on the perimeter of $\mathcal P$ with distance sum at most $\tfrac{Sk}{n}$: For the former step, at least one of the vertices of $\mathcal P$ must work; we will show that the sum of the distance sums of each vertex is at least $Sk$. Indeed, we can swap the order of summation – it is easy to check that the sum of the distances from any $P_i$ to the vertices of $\mathcal P$ is at least $S$, so the total sum is at least $Sk$. For the latter step, at least one of the midpoints of the sides of $\mathcal P$ must work; we will show that the sum of the distance sums of each midpoint is at most $Sk$. Once again, we swap the order of summation – this time, we verify that the sum of the distances from any $P_i$ to the midpoints of the sides of $\mathcal P$ is at most $S$, so the total sum is at most $Sk$.

This problem is strongly motivated by equality cases. I claim that $t = 30\left(1+\sqrt 5\right)$ works; in fact, it is the only value of $t$ that works. First, consider the four corners $Q_1, Q_2, Q_3, Q_4$. For any point $P$ on the exterior of the square, say on side $\overline{Q_1Q_2}$, we have \[PQ_1+PQ_2+PQ_3+PQ_4 = 1 + PQ_3+PQ_4 \geq 1 + \sqrt 5.\]Summing over all $120$ points $P$, there exists an index $i$ such that $A_1 Q_i + A_2Q_i + \cdots + A_{120} Q_i \geq 30\left(1+\sqrt 5\right)$. Similarly, consider the four midpoints $P_1, P_2, P_3, P_4$ of the edges. For any point $P$, we have \[PP_1+PP_2+PP_3+PP_4 \leq 1 + \sqrt 5\]which can be seen say by letting $P$ lie on the side with midpoint $P_1$ and setting $PP_1 = x$. It follows that there exists an index $j$ such that $A_1P_j + A_2P_j + \cdots + A_{120} P_j \leq 30\left(1+\sqrt 5\right)$. Thus, by intermediate value theorem, there exists a point $Q$ for which the sum is exactly $30\left(1+\sqrt 5\right)$, as needed. Remark: The answer could only conceivably be $30\left(1+\sqrt 5\right)$ because taking the $120$ points to be copies of the $P_i$ and $Q_i$ respectively yields two possible intervals for the sum which only intersect at $30\left(1+\sqrt 5\right)$.

Group Solved with kotmhn. First some notation: Let the vector $\mathbf P$ denote $\mathbf P=(P_1,P_2,...,P_{120})$ and let $d(\mathbf P,Q)=|P_1Q|+|P_2Q|+...+|P_{120} Q|$ Consider any two such $\mathbf P$ and $\mathbf P'$. If we can show that the range of possible values of $d(\mathbf P,Q)$ and $d(\mathbf P', Q')$ for some points $Q$ and $Q'$ always overlap, we would be done by continuity of $d(\mathbf P, Q)$. AFTSOC this is not the case: WLOG by IVT we have $d(\mathbf P,Q)> d(\mathbf P', Q')$ for all $Q, Q'$. This is clearly not true as we take $(Q,Q')=(P'_i, P_i)$, sum over all the inequalities that we get by varying $i$ to reach a contradction. $\blacksquare$. .