If I understand your question, it is why should the term "effectiverate" be used? This is a label referring to the interest rate per period in the problem, with that period being 1/2 a time interval. Interest rates are normally quoted on an annual basis which is useful for purposes of comparison, and are often referred to as the nominal annual rate. Given a nominal interest rate 0.04 the effective interest rate e is given by ((1+e)^2)-1 = 0.04 where there are 2 periods in one year. Sometimes the term effective rate is used by bankers to mean either the nominal rate or the effective rate as defined here, which can be confusing. The naive approach using "effectiverate" showed Daniel's required deposit was different because he used continuous compounding rather than effective rate. In making this type of calculation we implicitly assume an average rate of interest over the term of the investment; the answer obtained depends on which mean rate of return is used, which could be arithmetic mean rate of return, or geometric mean rate of return, the latter being assumed by "effectiverate". How much do you think Ludwig Schläfl would have invested if there were penalties for not delivering on payments? Why is your answer of 3378.77 different to the other two?

Assume that payments occur every 6 months, and payments and interest receipts coincide and take a naive approach

payments = {500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
(*made 6 monthly*)
Print["payments required = ", payments];
effectiverate = (1 + .04)^0.5 - 1;
(*compunding every 6 months*)
Print["effectiverate = ", effectiverate];
bondelements = Table[payments[[i]]/(1 + effectiverate)^i, {i, 1, Length[payments]}];
(*investment components to make payments*)
p = Apply[Plus, bondelements];
(*principal investment needed*)
Print["principal investment required = ", p];

output is

payments required = {500,673,245,345,245,674,425,145,37,423}

effectiverate = 0.0198039

principal investment required = 3381.8

Use of standard expressions by Daniel gives another answer

payments = {500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
tvm = TimeValue[Cashflow[payments, 1/2], EffectiveInterest[.04, 0], -1/2];
Print["principal investment required = ", tvm];

principal investment required = 3375.65

But this is reconciled if we use annual compounding in the code, EffectiveInterest[.04,1]

payments = {500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
tvm = TimeValue[Cashflow[payments, 1/2], EffectiveInterest[.04, 1], -1/2];
Print["principal investment required = ", tvm];

principal investment required = 3381.8

It is important to sketch a time-line diagram when timing or amounts become irregular, such as when interest is calculated annually on average balances. Using the Ruffle[list1,list2] command is useful for resolving irregularities in the problem formulation.

@Udo - The typo was a mistake on my part but even with it corrected it doesn't work.

Doesn't

In[13]:= Solve[Fold[1.02 #1 - #2 &, x, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}] == 0, x]
Out[13]= {{x -> 3378.77}}

help? (The interest is per year, the payment starts 6 month after the deposit and in the end no money remains.)

And this shows the account history just after a payment resp. the initial deposit

ListPlot[Transpose[
  {Range[0, 5, 0.5], FoldList[1.02 #1 - #2 &, 3378.77, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}]}
 ], PlotLabel -> "Tim pays the bill"]

and this is what has been payed every half year:

In[32]:= (1.02 Most[#] - Rest[#])&[FoldList[1.02 #1 - #2 &, 3378.77, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}]]
Out[32]= {500., 673., 245., 345., 245., 674., 425., 145., 37., 423.}

probe it

In[33]:= % - {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}
Out[33]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}