Imagine to have two urns (A and B) with equal number of balls (N) numbered from 1 to N. Each extraction consists of taking a ball from urn A and one from urn B, and then multiplying the respective numbers. Balls are discarded after and not reentered. Let's make N Extraction, and calculate the sum of the N resulting products.
I found the expected value for the sum being:
$$\frac{N(N+1)^2}{4}$$
which is a peace of the information that I'm looking for. Does anybody have a clue of what distribution is Sum going to follow. It's going to be close to a normal for big Ns, but not exactly since it's goint to have a well defined Min and a Max. What about the variance and standard deviation of this thing?
In this answer the route to finding variance, based on bilinearity of covariance and symmetry.
You can write the sum as $S=P_1+\cdots+P_N$ where the $P_i$ are the products.
Then: $$\mathsf{Var}(S)=\mathsf{Cov}(P_1+\cdots+P_N,P_1+\cdots+P_N)=N\mathsf{Var}(P_1)+N(N-1)\mathsf{Cov}(P_1,P_2)$$
Try to find $\mathsf{Var}(P_1)$ and $\mathsf{Cov}(P_1,P_2)$ yourself.
I have to go now, and later I will come back to have a second look.
edit (continuation)
Let $P_{i}=X_{i}Y_{i}$ where $X_{i}$ denotes the $i$-th number from urn A and $Y_{i}$ denotes the $i$-th number of urn B.
Then all $X_{i}$ and $Y_{i}$ for every $i\in\left\{ 1,\dots,N\right\} $ the random variables $X_{i}$ and $Y_{i}$ have uniform distribution over $\left\{ 1,\dots,N\right\} $.
Further the random vectors $\left(X_{1},\dots,X_{N}\right)$ and $\left(Y_{1},\dots,Y_{N}\right)$ are independent.
$$\mathbb{E}P_{1}=\mathbb{E}X_{1}Y_{1}=\mathbb{E}X_{1}\mathbb{E}Y_{1}=\left(\mathbb{E}X_{1}\right)^{2}$$
$$\mathbb{E}P_{1}^{2}=\mathbb{E}X_{1}^{2}Y_{1}^{2}=\mathbb{E}X_{1}^{2}\mathbb{E}Y_{1}^{2}=\left(\mathbb{E}X_{1}^{2}\right)^{2}$$
$$\mathsf{Var}P_{1}=\mathbb{E}P_{1}^{2}-\left(\mathbb{E}P_{1}\right)^{2}=\left(\mathbb{E}X_{1}^{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}$$
$$\mathsf{Covar}\left(P_{1},P_{2}\right)=\mathbb{E}P_{1}P_{2}-\mathbb{E}P_{1}\mathbb{E}P_{2}=\mathbb{E}X_{1}Y_{1}X_{2}Y_{2}-\left(\mathbb{E}X_{1}\right)^{4}=$$$$\mathbb{E}X_{1}X_{2}\mathbb{E}Y_{1}Y_{2}-\left(\mathbb{E}X_{1}\right)^{4}=\left(\mathbb{E}X_{1}X_{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}$$
$$\mathsf{Var}\left(S\right)=N\left[\left(\mathbb{E}X_{1}^{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}\right]+N\left(N-1\right)\left[\left(\mathbb{E}X_{1}X_{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}\right]=$$$$N\left(\mathbb{E}X_{1}^{2}\right)^{2}+N\left(N-1\right)\left(\mathbb{E}X_{1}X_{2}\right)^{2}-N^{2}\left(\mathbb{E}X_{1}\right)^{4}$$
What remains now is finding $\mathbb{E}X_{1}$, $\mathbb{E}X_{1}^{2}$ and $\mathbb{E}X_{1}X_{2}$. I leave that to you.