Product of 2-digit numbers 
31 people are dancing in a circle formation. Each people’s age is a 2-digit number, and for each digit we know that the units’ digit is equal to the tens’ digit of the person who is in the clockwise position, while the tens’ digit is equal to the units’ digit of the person who is in the counter-clockwise position. Any 2 neighboring digits are different. Check if the product of all 31 ages can be a perfect square.

I don’t know how to start :(
All I can tell is that the digits appear in pairs:
If we have, for example 31 digits, $d_1, d_2, \ldots, d_{31}$, then the ages are:
$$
10 \cdot d_1+d_2,
10 \cdot d_2+d_3,
\ldots,
10 \cdot d_{31}+d_1
$$
and obviously the product is
$(10 \cdot d_1+d_2)
\cdot (10 \cdot d_2+d_3)
\cdots (10 \cdot d_{31}+d_1)$.
But I don’t know how to continue.

It's possible. To restate the goal, we want a cycle of digits $d_1 \to d_2 \to d_3 \to \ldots \to d_{31} \to d_1$, such that
$$
(d_1 d_2) (d_2d_3) (d_3d_4) \ldots (d_{31} d_1)
$$
is a perfect square.
We can build this up by combining small cycles. First,
$$2 \to 7 \to 2$$
is a nice move, as it gives $27 \cdot 72$ -- dividing out squares, we have $3 \cdot 2 = 6$.
We can also do the move
$$
2 \to 5 \to 2
$$
which is $25 \cdot 52$, or just $13$ remaining.
And we also have
$$
2 \to 6 \to 9 \to 2
$$
which is $26 \cdot 69 \cdot 92$, or $2 \cdot 13 \cdot 3 \cdot 23 \cdot 23 \cdot 4$, or just $6 \cdot 13$.
So all of these together give a perfect square cycle of total length $2 + 2 + 3 = 7$:
$$
\underbrace{2 \to 7 \to 2 \to 5 \to 2 \to 6 \to 9 \to 2}_{\text{perfect square cycle}}. \tag{a}
$$
But that's not good enough -- we need another way of getting a perfect square, because cycles of length $7$ can't add up to get total length $31$.
So here's another cycle:
$$
\underbrace{2 \to 4 \to 2 \to 1 \to 2}_{\text{perfect square cycle}}. \tag{b}
$$
Just to verify this, we have $24 \cdot 42 \cdot 21 \cdot 12$, dividing out squares and factoring $6 \cdot (6 \cdot 7) \cdot (3 \cdot 7) \cdot 3$, so everything ends up squared. This time, the cycle has length $4$.
(This was a bit more complicated than necessary. We could just have done $2 \to 5 \to 2 \to 5 \to 2$, or $2 \to d \to 2 \to d \to 2$ for any $d$, for that matter.)
Then we are done: by starting from $2$ and traversing cycle (a) one and cycle (b) six times, the total length is
$$
7 + 4 + 4 + 4 + 4 + 4 + 4 = 31.
$$
Some intuition (if you know graph theory)
The whole length $31$ cycle can be thought of as a cyclic path in the complete graph with $9$ vertices, where each vertex is a different digit. The path of length $31$ then decomposes into a bunch of cycles of small length ($\le 9$). The procedure to decompose into small cycles is as follows: for any cycle of length greater than $9$, it must repeat some vertex twice; therefore, it splits into two cycles from the repeated vertex.
So the goal is basically to find a bunch of small cycles whose total length is $31$, and which multiply to a perfect square. Hence, my approach was to start looking for small cycles and what they multiply to, and then try to get them to cancel out.
