Iranian Mathematical Olympiad

The answer is $I = (1, 1 + \frac{1}{d})$.

Assume that the desired interval is of the form $(b, c)$. For some $q, r$ in $(b, c)$, put $a_i = q$ for odd and $a_i = r$ for even $i$, where $0 \le i \le 2d - 1$. Then
$$P(-1) = 1 - dq + dr.$$
Since $P(x)$ has no real root, we must have $P(-1) > 0$. Thus, $q - r < \frac{1}{d}$. We can assume that $c - b = \frac{1}{d}$. Therefore, consider the interval as $(b, b + \frac{1}{d})$. It is easy to find that $b > 0$, since by choosing $a_0 = b + \epsilon$ should be positive. Now, assign the following numbers to the coefficients of the polynomial.
$$a_{2d-1} = a_{2d-3} = \dots = a_1 = b + \frac{1}{d} + \epsilon, \quad a_{2d} = a_{2d-2} = \dots = a_0 = b + \epsilon,$$
for some sufficiently small $\epsilon > 0$.
It is clear that for all positive real $x$, $P(x) > 0$. For all negative real $x$, putting $x = -t$, where $t > 0$, then,
$$P(-t) = t^{2d} - \left(b + \frac{1}{d}\right) t^{2d-1} + bt^{2d-2} - \dots - \left(b + \frac{1}{d}\right) t + b + \epsilon Q(t),$$
for some polynomial $Q(t)$, $\deg Q(t) = 2d - 1$. As $\epsilon$ tends to zero, it remains to find all $b > 0$ such that
$$R(t) = t^{2d} - \left(b + \frac{1}{d}\right) t^{2d-1} + b t^{2d-2} - \cdots - \left(b + \frac{1}{d}\right) t + b \geq 0.$$
Note that $R(1) = 0$. Therefore, $R'(1)$ should be zero. That is,
$$2d - (2d - 1) \left(b + \frac{1}{d}\right) + (2d - 2)b - \cdots - \left(b + \frac{1}{d}\right) = 0.$$
Hence, we claim the interval $I = (1, 1 + \frac{1}{d})$ works! It is obvious that $P(x)$ has no positive real roots. Now, we prove that $P(x) > 0$ for all negative real $x$. Put $x = -t, t > 0$.
Then,
$$P(-t) > t^{2d} - \left(1 + \frac{1}{d}\right) t^{2d-1} + t^{2d-2} - \dots - \left(1 + \frac{1}{d}\right) t + 1.$$
Thus it remains to prove that
$$\frac{t^{2d} + t^{2d-2} + \dots + t^2 + 1}{d+1} \ge \frac{t^{2d-1} + t^{2d-3} + \dots + t^3 + t}{d}.$$
Now, since $t^{2d} + 1 \geq t^{2d-2k+1} + t^{2k-1}$ and $t^{2k} + t^{2k-2} \geq 2t^{2k-1}$, we are done.
Thus,
$$b \le \frac{1}{d} \lim_{t \to 1^+} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t - 1}.$$
and
$$b \geq \frac{1}{d} \lim_{t \to 1^-} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t - 1},$$
since the function
$$\frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t - 1}$$
has a limit as $t$ approaches 1. Thus, we find that
$$\lim_{t \to 1^+} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t-1} = \lim_{t \to 1^-} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t-1}.$$
Hence,
$$b = \frac{1}{d} \lim_{t \to 1} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t - 1}.$$
Finally note that
$$\begin{align*} & \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t-1} \\ &= \frac{(t^{2d} - t^{2d-1}) + (t^{2d} - t^{2d-3}) + \dots + (t^{2d} - t)}{(t-1)(t^{2d-2} + t^{2d-4} + \dots + 1)} \\ &= \frac{(t-1)(t^{2d-1} + t^{2d-3}(t^2+t+1) + \dots + t(t^{2d-2} + \dots + 1))}{(t-1)(t^{2d-2} + t^{2d-4} + \dots + 1)} \end{align*}$$
Thus,
$$\lim_{t \to 1} \frac{dt^{2d} - (t^{2d-1} + t^{2d-3} + \dots + t)}{t^{2d-1} - t^{2d-2} + \dots + t - 1}$$
$$= \lim_{t \to 1} \frac{t^{2d-1} + t^{2d-3} (t^2 + t + 1) + \dots + t (t^{2d-2} + \dots + 1)}{t^{2d-2} + t^{2d-4} + \dots + 1} = \frac{d^2}{d} = 1.$$
That is, $b = \frac{1}{d}$ and $d = 1$.