Let the amount borrowed be P in present-day values. The first 4 payments will be paid at the beginning of years 1 to 4. The remaining 5 payments will be at the end of years 5 to 9. Let's say year 1 is 2016 and the first payment is January, then year 4 is 2019; year 5 is 2020 and year 9 is 2024. The money is totally repaid in December 2024.
After the first payment the amount remaining to be paid will be P-2000. Before the second payment, this residue will appreciate to 1.0685(P-2000)=1.0685P-2000*1.0685.
The 2nd payment will reduce this to 1.0685P-2000*1.0685-2000;
the 3rd payment to 1.0685^2P-2000*1.0685^2-2000*1.0685-2000;
and the 4th to 1.0685^3P-2000*1.0685^3-2000*1.0685^2-2000*1.0685-2000.
Rewrite this: 1.0685^3P-2000(1.0685^3+1.0685^2+1.0685+1)
The geometric series in brackets we can call S. 1.0685S=1.0685^4+...+1.0685.
Subtract S and we have: 0.0685S=1.0685^4-1, so S=(1.0685^4-1)/0.0685=4.43
After the fourth payment the amount remaining is 1.0685^3P-2000*4.43=1.0685^3P-8860.18 approx.
This amount has to be paid off using the remaining 5 payments of R5000.
The 4th payment is made at the beginning of year 4 (e.g., 2019), but the 5th payment is made at the end of the 5th year (2020), so the balance to be paid accrues over the remainder of the 4th year and practically the whole of the 5th year (e.g., December 2020). That is slightly less than two years. We'll take this as being two years, so in the "waiting period" of 2 years, the remainder accrues 2 years' value: 1.0685^2(1.0685^3P-8860.18) and then the 5th payment is made leaving 1.0685^2(1.0685^3P-8860.18)-5000. Let's call the first term in this expression p, so p-5000 remains after the 5th payment. This is exactly analagous to the way we started, so after 5 years we have 1.0685^4p-5000(1.0685^5-1)/0.0685=1.0685^4p-5000*5.7336=1.0685^4p-28667.76.
The debt is paid off so 1.0685^4p=28667.76 and p=R21993.56.
1.0685^2(1.0685^3P-8860.18)-5000=p=R21993.56. From this P=R26644.56.