Suppose for contradiction that all 101 polynomials have all real roots. Let $P_k(x)$ denote the $k$-th polynomial:
$$P_k(x) = a_{100+k}x^{100} + 100a_{99+k}x^{99} + a_{98+k}x^{98} + a_{97+k}x^{97} + \dots + a_{2+k}x^2 + a_{1+k}x + a_k,$$
where all indices are modulo 101.

Note that all coefficients $a_i$ are positive reals.

Let $S$ be the $101 \times 101$ matrix whose $k$-th row (for $k = 0, 1, \dots, 100$) is the vector of coefficients of $P_k(x)$, i.e.,
$$(a_{100+k},\ 100a_{99+k},\ a_{98+k},\ a_{97+k},\ \dots,\ a_{2+k},\ a_{1+k},\ a_k)$$
with indices modulo 101.

Let us consider the sum of all these polynomials:
$$\sum_{k=0}^{100} P_k(x) = \sum_{k=0}^{100} \left( a_{100+k}x^{100} + 100a_{99+k}x^{99} + a_{98+k}x^{98} + \dots + a_{1+k}x + a_k \right)$$
Let us compute the coefficient of $x^m$ in this sum for $m = 0, 1, \dots, 100$.

- For $m = 100$, the coefficient is $\sum_{k=0}^{100} a_{100+k} = \sum_{k=0}^{100} a_k$ (since $100+k$ runs over all residues mod 101 as $k$ runs from 0 to 100).
- For $m = 99$, the coefficient is $\sum_{k=0}^{100} 100a_{99+k} = 100 \sum_{k=0}^{100} a_k$.
- For $m = 98$, the coefficient is $\sum_{k=0}^{100} a_{98+k} = \sum_{k=0}^{100} a_k$.
- For $m = 97$, the coefficient is $\sum_{k=0}^{100} a_{97+k} = \sum_{k=0}^{100} a_k$.
- ...
- For $m = 1$, the coefficient is $\sum_{k=0}^{100} a_{1+k} = \sum_{k=0}^{100} a_k$.
- For $m = 0$, the coefficient is $\sum_{k=0}^{100} a_k$.

So, the sum is:
$$\sum_{k=0}^{100} P_k(x) = (\sum_{k=0}^{100} a_k) x^{100} + 100(\sum_{k=0}^{100} a_k) x^{99} + (\sum_{k=0}^{100} a_k) x^{98} + \dots + (\sum_{k=0}^{100} a_k) x^0$$
That is,
$$\sum_{k=0}^{100} P_k(x) = (\sum_{k=0}^{100} a_k) \left( x^{100} + 100x^{99} + x^{98} + x^{97} + \dots + x^2 + x + 1 \right)$$

Now, $\sum_{k=0}^{100} a_k > 0$ since all $a_k > 0$.

Therefore, the sum of all 101 polynomials is a positive constant times the polynomial
$$Q(x) = x^{100} + 100x^{99} + x^{98} + x^{97} + \dots + x^2 + x + 1$$

But $Q(x)$ is a degree 100 polynomial with real coefficients. Let us check whether $Q(x)$ can have all real roots.

Suppose $Q(x)$ has all real roots. But the coefficient of $x^{99}$ is $100$, which is much larger than the other coefficients (which are all $1$). For a real-rooted degree 100 polynomial with all positive coefficients, the sequence of coefficients must be log-concave (by Newton's inequalities). That is, for all $i$, $a_i^2 \geq a_{i-1} a_{i+1}$.

But for $x^{99}$, the coefficient is $100$, and for $x^{100}$ and $x^{98}$, the coefficients are $1$ each. So,
$$100^2 = 10,000 \geq 1 \cdot 1 = 1$$
which is true, but let's check the log-concavity at $x^{99}$:
$$100^2 \geq 1 \cdot 1$$
which is true, but let's check at $x^{98}$:
$$1^2 \geq 100 \cdot 1 \implies 1 \geq 100$$
which is false.

Therefore, the sequence of coefficients is not log-concave, so $Q(x)$ cannot have all real roots.

But $Q(x)$ is a sum of polynomials, each of which (by assumption) has all real roots. The sum of real-rooted polynomials with positive coefficients is also real-rooted only if the sequence of coefficients is log-concave. Since $Q(x)$ is not real-rooted, at least one of the $P_k(x)$ cannot be real-rooted.

Therefore, it is impossible that all 101 polynomials have all roots real.