Machine learning technique to calculate weighted average weights?
Yes it is definitely possible to calculate optimised weightings provided you have some training examples where you know the document fields, the query, and either the outcome (relevant/not-relevant) or the desired score. 

I think your training feature set should be the query score in range [0.0,1.0] for each field of each example. The training label should be either relevance 0 or 1 for each example, or the relevance score that the example has.

If you have a target score for each example

You want to determine the weights $W_i$ to use for each field $i$. Your calculated relevance score would be $\hat{y} = \sum_{i=1}^{N_{fields}} W_i * X_i$ where the caret signifies this is the estimate from your function and $N_{fields}$ is the number of field. Note I am ignoring your original idea of dividing by the sum of all $W_i$, because it makes things more complex. You can either add that term or force the sum to be equal to 1.0 if you wish (I am not going to show you how though, as this answer would get too long, and it probably won't help you much)

With a target score and training data, the simplest approach is to find the weights which cause the lowest error when used with the training data. This is a very common goal in supervised learning. You need a loss function. Having a target scalar value means you can use difference from target and a very common loss function for this kind of regression problem is the mean squared error:

$$E = \frac{1}{N_{examples}} \sum_{j=1}^{N_{examples}} (\hat{y}_j - y_j)^2$$

Where $\hat{y}_j$ is your calculated relevance score for example $j$ and $y_j$ is your training label for the same example.

There are a few different ways to solve for lowest $E$ with this loss function, and it is one of the simplest to solve. If you express your weights as a vector $W$ length $N_{fields}$ your example features as a matrix $X$ size $N_{examples} \times N_{fields}$ and the labels as a vector $Y$ length $N_{examples}$ then you can get an exact solution to minimise loss using the linear least squares equation

$$W = (X^TX)^{-1}X^TY$$

There are other approaches that work too - gradient descent or other function optimisers. You can look these up and see which you would prefer to use for your problem. Most programming languages will have a library with this already implemented.

Note that you will likely get scores greater than 1.0 or less than 0.0 from some document/query pairs.

You will have to use a adjust the technique if you want to divide by total of all weights or want sum of all weights equal to 1 in your scoring system.

If you have a relevance 0 or 1 for each example

You have a classification problem, relevant or not are your two classes. This can still be made to work, but you will want to change how you calculate your weighted score and use logistic regression. 

Your weighted score under logistic regression would be:

$$\hat{y} = \frac{1}{1 + e^{-(b + \sum_{i=1}^{N_{fields}} W_i * X_i)}}$$ 

Where $b$ is a bias term. This looks complicated, but really it is just the same as before but mapped by a sigmoid function to better represent class probabilities - the result is always between 0 and 1.

You can look up solvers for logistic regression, and most stats or ML libraries will have functions ready to use.

Caveats

You have made a starting assumption that a simple combined relevance score will lead to a useful end result for your users performing search. This has led to simple linear models looking like a good solution. However, this may not be the case in practice, and you may need to re-visit that assumption.