We enumerate the children by numbers $1, 2, \dots, 20$. Let us denote by $d_i$ the number of ribbons that are being held by the child with number $i$. Since each ribbon is being held by two children we have that
$$\sum_{i=1}^{20} d_i = 2 \cdot 100 = 200.$$
The number of all pairs of ribbons is $\frac{100 \cdot 99}{2} = 4950$.
A pair of ribbons such that its four ends are not being held by different children is given by a pair of ribbons such that there is a child who holds one end of each of the two ribbons in that pair.
Hence
$$\sum_{i=1}^{20} \frac{d_i \cdot (d_i - 1)}{2} = 4950 - 4050 = 900.$$
From here it follows that
$$\sum_{i=1}^{20} d_i^2 = 1800 + \sum_{i=1}^{20} d_i = 1800 + 200 = 2000.$$
By the AM–QM inequality we have
$$10 = \frac{d_1 + d_2 + \dots + d_{20}}{20} \le \sqrt{\frac{d_1^2 + d_2^2 + \dots + d_{20}^2}{20}} = 10.$$
Since the equality in AM–QM inequality is obtained if and only if all $d_i$ are equal, the claim follows.