I can see many possible approaches, but let me outline one approach that I think might be effective and might give reasonable accuracy with a reasonable amount of effort.  It uses Monte Carlo simulation, independence, and linearity of expectation.  I'm going to break it down into bite-sized pieces, by identifying a set of smaller subproblems and explaining how to solve each subproblem, then showing how to combine those solutions to the subproblems to solve the original programming contest problem.

**Subproblem 1.** Given two circles $C,C'$ in the square, determine whether they intersect.

**Solution 1.** Let $C$ be centered at the point $(x,y)$ and have radius $r$, and $C'$ be centered at $(x',y')$ with radius $r'$.  Note that the two circles overlap if and only if the distance between the two centers is at most $r+r'$, i.e., if and only if $(x-x')^2 + (y-y')^2 \le (r+r')^2$.  This condition is easy to test, as a function of $x,y,r,x',y',r'$.

**Subproblem 2.** Suppose the first circle is given by $C$.  Compute the probability $p(C)$ that the second circle will not intersect $C$, as a function of $C$.

**Solution 2.**  We will use Monte Carlo simulation.  In particular, perform 10,000 trials, where in each trial, we randomly drop the second circle $C'$ and test whether $C,C'$ intersect (using solution 1 above to test for an intersection).  Count the number of trials where they do not intersect and divide by 10,000; that is your estimate of $p(C)$.  (It may be possible to solve this subproblem analytically, by solving a three-dimensional integral; however, that does not sound fun.  Monte Carlo simulation is easier.)

**Subproblem 3.** Suppose the first circle is given by $C$.  Compute the probability $p'(C)$ that none of the remaining $N-1$ circles will intersect $C$, as a function of $C$.

**Solution 3.** Note that, by independence, $p'(C) = p(C) \times \cdots \times p(C) = p(C)^{N-1}$.  This is easy to compute.

**Subproblem 4.** Compute the probability $q_1$ that, if you drop $N$ circles randomly, none of the last $N-1$ circles has any intersection with the first circle.

**Solution 4.** Note that, if we let $C$ be a random variable, we have $q_1 = E[p'(C)]$, where the expectation is taken over $C$.  Use Monte Carlo simulation.  Perform 10,000 trials, where in each trial you randomly choose a circle $C$, then you compute $p'(C)$ (using solution 3).  Average the results over all of the trials.  That is your estimate of the probability $q_1$.

**Subproblem 5.** Compute the probability $q_i$ that, if you drop $N$ circles randomly, the $i$th circle has no intersection with any of the other $N-1$ circles.

**Solution 5.** By symmetry, $q_i = q_1$, so solution 4 already provides the answer.

**The original problem.** Compute the expected number of circles that have no intersection with any other circle, if you drop $N$ circles randomly.

**Solution.** By linearity of expectation, this is $q_1+q_2+\cdots+q_N$.  By solution 5, this is $N \times q_1$.  Now use solution 4 to compute $q_1$, and you have your answer to the original problem.