// DO NOT DO THIS. It is quadratic-time!
StringBuilder sb = new StringBuilder();
for(int i=0;i<100;i++){
    sb.insert(0, Integer.toString(i));
}

Warning: It defeats the purpose of StringBuilder, but it does what you asked.

Better technique (although still not ideal):

1. Reverse each string you want to insert.
2. Append each string to a StringBuilder.
3. Reverse the entire StringBuilder when you're done.

This will turn an O(n²) solution into O(n).

just do a .reverse on the sb.toString() at the end.. surely that's 10x more readable

Yes, reversing the enumerable is much faster.

For example:

var numStrings = 80000;
var strings = new List<String>();
for(var i = 0; i < numStrings; i++)
{
    strings.Add(Guid.NewGuid().ToString());
}

var sw = new Stopwatch();
sw.Start();
var sb = new StringBuilder();
foreach(var str in Enumerable.Reverse(strings))
    sb.Append(str);

sw.Stop();
sw.ElapsedMilliseconds.Dump(); // 13 milliseconds
sb.Dump();

sw = new Stopwatch();
sw.Start();
sb = new StringBuilder();
foreach(var str in strings)
    sb.Insert(0, str);

sw.Stop();
sw.ElapsedMilliseconds.Dump(); // 42063 milliseconds
sb.Dump();