Dear Fellow,
You can't move $E$ (!), hence there is no contradiction with it having self-intersection -1.
Indeed, if you take a normal vector field along $E$, it will necessarily have degree -1 (i.e.
the total number of poles is one more than the total number of zeroes), or (equivalently),
the normal bundle to $E$ in the blown-up surface is $\mathcal O(-1)$.
[Added:] Here is a version of the argument given in David Speyer's answer, which is rigorous modulo basic facts about intersection theory:
Choose two smooth very ample curves $C_1$ and $C_2$ passing through the point $P$ being blown-up in different tangent directions. (We can construct these using hyperplane sections in some projective embedding, using Bertini; smoothness is just because I want $P$ to be a simple point on each of them.) If the $C_i$ meet in $n$ points away from $P$, then
$C_1\cdot C_2 = n+1$.
Now pull-back the $C_i$ to curves $D_i$ in the blow-up. We have
$D_1 \cdot D_2 = n + 1$. Now because $C_i$ passes through $P$, each $D_i$ has the form
$D_i = D_i' + E,$ where $D_i'$ is the proper transform of $C_i$, and passes through $E$ in a single point (corresponding to the tangent direction along which $C_i$ passed through $P$).
Thus $D_1'\cdot D_2' = n$ (away from $P$, nothing has changed, but at $P$, we have separated
the curves $C_1$ and $C_2$ via our blow-up).
Now compute $n+1 = D_1\cdot D_2 = D_1'\cdot D_2' + D_1'\cdot E + E\cdot D_2' + E\cdot E
= n + 1 + 1 +
E\cdot E$, showing that $E\cdot E = -1$. (As is often done, we compute the intersection of curves that we can't move into a proper intersection by adding enough extra stuff that we can compute the resulting intersection by moving the curves into proper position.)
Your question is actually far too general, so let me assume $n=2$. Also in this case, there are only partial results.
In the case where all $a_i$ are equal to $1$, Kurchle and (independently) Xu showed that
$$H=\pi^*(dL) - \sum_{i=1}^r E_i$$
(where $L$ is the class of a line) is ample, provided that $H^2 > 0$ and $d \geq 3$.
Later, Szemberg and Tutaj-Gasinska, in their paper General blow-ups of the projective plane (Proceedings of the Amer. Math. Soc. 130, 2002), proved the following (non optimal) result:
Theorem. Let $X$ the blow-up of $\mathbb{P}^2$ at $r$ general points and let $k\geq 2$ and $r$ be integers such that $d \geq 3k+1$. If $r \leq \frac{d^2}{k^2} -1$, then the line bundle $$H=\pi^*(dL) -k \sum_{i=1}^r E_i$$ is ample.
I refer you to Szemberg-Tutaj-Gasinska paper for more details on this problem and on its relations with the so-called Nagata Conjecture.
In higher dimensions, there is a paper by Angelini that generalizes the result of Xu for $n=3$, in the case where all blown-up subvarieties are points. See Ample divisors on the blow up of $\mathbb{P}^3$ at points , Manuscripta Mathematica 93 (1997).
Best Answer
Let $[x:y:z:w]$ be the homogeneous coordinates on $\mathbb{P}^3$ and suppose that the point blown up is given by the vanishing of the first three coordinates. Introduce a projective plane with coordinates $[X:Y:Z]$ then the blowup is a closed subset in $\mathbb{P}^3 \times \mathbb{P}^2$ given by the equations $$ xY = yX; xZ = zX; yZ = zY. $$ The bundle $\mathcal{O}(1) \boxtimes \mathcal{O}(1)$ on $\mathbb{P}^3 \times \mathbb{P}^2$gives the Segre Embedding, so it is very ample on the product of projective spaces. Its restriction to the blowup is also very ample and it must be of the form $dH - k E$. We can computed $d$ and $k$ by intersecting with (a) a line in $\mathbb{P}^3$ that avoids the point blown up, which gives $d = 2$; and (b) with a line in the exceptional fiber $[0:0:0:1] \times \mathbb{P}^2$, which gives $k=1$. So $2H - E$ is very ample.